The number of moths in the collection is given as follows:E) 84.
How to obtain the number of moths?The number of moths in the collection is obtained by applying the proportions in the context of the problem.
The total number of insects in the collection is given as follows:112 insects.
The fraction relative to moths in the collection is given as follows:3/4.
Hence the number of moths in the collection is given as follows:3/4 x 112 = 84.
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The question in English :
Luis's insect collection is
composed of 112 insects, and 3/4 of them
they are butterflies. How many butterflies are in the
collection?
(A) 28
(B) 37
(C) 64
(D) 75
(E) 84
What is the partial fraction decomposition of startfraction 7 x squared minus 6 x + 9 over 3 x (4 x squared + 9) endfraction?
Answer:
A:(1/3x)+(x-2/4x^2+9)
Step-by-step explanation:
Edg2022
How many triangles are there?
Answer:
20
Steps-by-step explanation:
sorry mistake
Answer:
12
Explanation: I just counted them! :) and there was 6 in the middle and on the outside so 6 + 6 = 12! :)
1. The midpoint of KL is M(5, -7). One endpoint is K(9,-7). Find the coordinates of the other endpoint L.
O
O
O
(7.-7)
(1.-7)
(1,0)
Answer:
L (1, - 7 )
Step-by-step explanation:
give endpoints (x₁, y₁ ( and (x₂, y₂ ) then the midpoint is
( [tex]\frac{x_{1}+x_{2} }{2}[/tex] , [tex]\frac{y_{1}+y_{2} }{2}[/tex] )
here (x₁, y₁ ) = K (9, - 7 ) and (x₂, y₂ ) = L (x, y )
use the midpoint formula and equate to corresponding coordinates of M
[tex]\frac{9+x}{2}[/tex] = 5 ( multiply both sides by 2 to clear the fraction )
9 + x = 10 ( subtract 9 from both sides )
x = 1
and
[tex]\frac{-7+y}{2}[/tex] = - 7 ( multiply both sides by 2 )
- 7 + y = - 14 ( add 7 to both sides )
y = - 7
Then L = (1, - 7 )
Simplify the expression. 3^–9 • 3^6 • 3^6
Answer:
27
Step-by-step explanation:
(3^−9)(3^6)(3^6)
=1/19683(3^6)(3^6)
=1/19683(3^6)(3^6)
=(1/19683(729))(3^6)
=1/27(3^6)
=1/27(729)
=27
Answer:
27^3
Step-by-step explanation:
The following table shows the weights of nine subjects before and after following a particular diet for two months. Test the claim that the diet is effective in helping people lose weight Subject A B DEFGHI Before 168 180 157 132 202 124 190 210 171 After 162 178 145 125 171 126 180 195 163
Using the t-distribution, it is found that there is not enough evidence that the diet is helping people lose weight.
What are the hypothesis tested?At the null hypothesis, it is tested if the mean hasn't decreased, that is:
[tex]H_0: \mu_B \leq \mu_A[/tex]
[tex]H_0: \mu_B - \mu_A \leq 0[qtex]
At the alternative hypothesis, it is tested if the mean has decreased, that is:
[tex]H_1: \mu_B - \mu_A > 0[qtex]
What are the mean and the standard error for the distribution of differences?For each sample, they are given as follows:
[tex]\mu_B = 170.44, s_B = \frac{29.275}{\sqrt{9}} = 9.7583[/tex].[tex]\mu_A = 160.56, s_A = \frac{24.203}{\sqrt{9}} = 8.067[/tex].Hence, for the distribution of differences, they are:
[tex]\overline{x} = \mu_B - \mu_A = 170.44 - 160.56 = 9.88[/tex].[tex]s = \sqrt{s_B^2 + s_A^2} = \sqrt{9.7583^2 + 8.067^2} = 11.66[/tex]What is the test statistic and the decision?
The test statistic is:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.
Hence:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
[tex]t = \frac{9.88 - 0}{11.66}[/tex]
t = 0.85.
Considering a right-tailed test with 9 + 9 - 2 = 16 df, with a standard significance level of 0.05, the critical value is t = 1.7459. Since t = 0.85 < 0.85, there is not enough evidence that the diet is helping people lose weight.
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We can learn a lot about a population if we select a ______ of it. Group of answer choices population subset data set case
We can learn a lot about a population if we select a subset of it.
What is a subset?One kind of set is a sample space. It is a clear listing of every event that could occur in a statistical experiment. A statistical experiment's events are a subset of the sample space.
A subset is a smaller group of results that are part of the bigger group.
Subsets are events, and events are subsets. A subset is an event of a sample space, and an event is a potential result of an experiment. A random experiment's sample space is a set (S) that contains all of the experiment's potential outcomes.
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Which best describes how to Sovle the equation below?
26x = 74
A. Multiply both sides by 74.
B. Multiply both sides by 26.
C. Divide both sides by 26.
D. Divide both sides by 74.
Factor the greatest common factor: 28a3b4 20a2b2 − 16ab3. 4ab(7a2b 5a − 4b2) 4ab2(7a2b 5a − 4b) 4ab2(7a2b2 5a − 4b) 4ab(7a2b3 5a − 4b)
Answer:
4ab2(7a2b2 5a - 4b)
Step-by-step explanation:
28a3b4 20a2b2 − 16ab3
The GCF of 28, 20 and 16 is 4.
GCF of the variables is ab2
So the answer is
4ab2(7a2b2 5a - 4b)
Answer:
4ab2(7a2b2 5a - 4b)
Step-by-step explanation:
Assume a student received the following grades for the semester: History, B; Statistics, A; Spanish, C; and English, C. History and English are 6 credit-hour courses, Statistics is a 8 credit-hour course, and Spanish is a 6 credit-hour course. If 4 grade points are assigned for an A, 3 for a B, and 2 for a C, what is the weighted mean grade for the semester
The weighted mean grade for the semester is 8.22
What is mean?It is calculated by dividing the total number of values in a set of data, such as measurements or numbers, by the total number of values.
No. of credits for Spanish, History and English = 6
No. of credits for Statistics = 8
Grade points for A = 4
Grade points for B = 3
Grade points for C = 2
Weighted mean = (6 * 2 + 6 * 3 + 8 * 4 + 6 * 2) / 9
Weighted mean = (12 + 18 + 32 + 12) / 9
Weighted mean = 74 / 9 = 8.22
Hence, the weighted mean grade for the semester is 8.22.
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HELPPPP PLSSSSSSSS
-----
Answer:
Translation of 3 units to the left.
Vertical stretch by a factor of 2.
Translation of 5 units down.
Step-by-step explanation:
Transformations
For a > 0
[tex]f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}[/tex]
[tex]y=a\:f(x) \implies f(x) \: \textsf{stretched parallel to the y-axis (vertically) by a factor of}\:a[/tex]
[tex]f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}[/tex]
Parent function:
[tex]y=x^2[/tex]
Translate 3 units left
Add 3 to the variable of the function
[tex]\implies y=(x+3)^2[/tex]
Stretch vertically by a factor of 2
Multiply the whole function by 2:
[tex]\implies y=2(x+3)^2[/tex]
Translate 5 units down
Subtract 5 from the whole function:
[tex]\implies y=2(x+3)^2-5[/tex]
Please see the attached graphs for the final transformed function (as well as the graphed steps).
Answer:
a) vertical expansion by a factor of 2; translation 3 units left and 5 units down
b) see attached
Step-by-step explanation:
a.Describing transformations is all about matching patterns. The elements of the transformed function are matched with the elements of a transformation.
Vertical scalingA function is scaled vertically by multiplying each function value by some scale factor. In generic terms, the function f(x) is scaled vertically by the factor 'c' in this way:
original function: f(x)scaled by a factor of 'c': c·f(x)If we want the function f(x) = x² scaled vertically by a factor of 2, then we have
f(x) = x² . . . . . . . original function
2·f(x) = 2x² . . . . scaled vertically by a factor of 2
On a graph, each point is vertically twice as far vertically from some reference point (the vertex, for example) as it is in the original function graph.
Horizontal translationA function is translated to the right by 'h' units when x is replaced by (x -h).
original function: f(x)translated h units right: f(x -h)If we want the function f(x) = x² translated right by 3 units, we will have ...
f(x) = x² . . . . . . . . . . . original function
f(x -3) = (x -3)² . . . . . .translated right 3 units
Note that translation left by 3 units would give ...
f(x -(-3)) = f(x +3) = (x +3)² . . . . translated left 3 units
On a graph, each point of the left-translated function is 3 units left of where it was on the original function graph.
Vertical translationA function is translated upward by 'k' units when k is added to the function value.
original function: f(x)translated k units up: f(x) +kThe value of k will be negative for a translation downward.
If we want the function f(x) = x² translated down by 5 units, we will have ...
f(x) = x² . . . . . . . . . . . original function
f(x) = x² -5 . . . . . . . . .translated down 5 units
Combined transformationsUsing all of these transformations at once, we have ...
f(x) = x² . . . . . . . . . . . . . . . . . original function
c·f(x -h) +k = c·(x -h)² +k . . . scaled by 'c', translated h right and k up
Compare this to the given function:
y = 2(x +3)² -5
and we can see that ...
c = 2 . . . . . . vertical scaling by a factor of 2h = -3 . . . . . translation 3 units leftk = -5 . . . . . translation 5 units downThis is the pattern matching that is described at the beginning.
__
b.When graphing a transformed function, it is often useful to start with a distinctive feature and work from there. The vertex of a parabola is one such distinctive feature.
TranslationThe transformations move the vertex 3 units left and 5 units down from its original position at (0, 0). The location of the vertex on the transformed function graph will be at (x, y) = (-3, -5).
Vertical scalingThe graph of the parent function parabola (y= x²) goes up from the vertex by the square of the number of units right or left. That is, 1 unit right or left of the vertex, the graph is 1 unit above the vertex. 2 units right or left, the graph is 2² = 4 units above the vertex.
The scaled graph will have these vertical distances multiplied by 2:
±1 unit horizontally ⇒ 2·1² = 2 units vertically; points (-4, -3), (-2, -3)±2 units horizontally ⇒ 2·2² = 8 units vertically; points (-5, 3), (-1, 3)The graph of the transformed function is shown in blue in the attachment.
__
Additional comment
The vertical scale factor 'c' may have any non-zero value, positive or negative, greater than 1 or less than 1. When the magnitude is less than 1, the scaling is a compression, rather than an expansion. When the sign is negative, the graph is also reflected across the x-axis, before everything else.
Find the missing length of the triangle (Side AC)
Answer: 17.3 cm
Step-by-step explanation:
[tex]\tan 68^{\circ}=\frac{AC}{7}\\\\AC=7\tan 68^{\circ}\\\\AC \approx 17.3[/tex]
The volume of a cone is 37x³ cubic units and its height is x units.
Which expression represents the radius of the cone's base, in units?
O 3x
0 6x
о 3лх2
0 90х2
Answer: 6x
Step-by-step explanation:
The expression represents the radius of the cone's base is √111/π x = r
What is Three dimensional shape?a three dimensional shape can be defined as a solid figure or an object or shape that has three dimensions—length, width, and height.
The formula for the volume of a cone is V = (1/3)πr²h, where V is the volume,
r is the radius of the base,
h is the height, and π is a constant approximately equal to 3.14.
We are given that the volume of the cone is 37x³ cubic units and the height is x units.
Using the formula for the volume of a cone, we can write:
37x³ = (1/3)πr²x
37x² = πr²/3
37×3 x²/π = r²
111x²/π = r²
Take square root on both sides
√111/π x = r
Hence, the expression represents the radius of the cone's base is √111/π x = r
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what is the solution to the equation below? square root 2-3x/square root 4x=2
Answer:
X=4 easy blah blah blah it’s 4. jdjjdjfnfvbdnnfn
Determine the area of the figure.
Answer: (x + 11) cm²
Step-by-step explanation:
The formula for the area of a trapezoid is [tex]\frac{1}{2}(a+b)h[/tex], where
a and b are the lengths of the top and bottom sides, andh is the height of the figure.For this figure, [tex]a = 5[/tex] and [tex]b= x+6[/tex] (the top and bottom sides), while [tex]h=2[/tex] (the height).
If we suppose [tex]A[/tex] is the area, plugging these values in, we get
[tex]A = \frac{1}{2}(5 + x + 6)*2[/tex]
[tex]A = \frac{1}{2} * 2 * (x + 11)[/tex]
[tex]A = x + 11[/tex]
All the lengths are in cm, so the area will be (x + 11) cm².
When you roll two number cubes, what are the odds, in simplest form, in favor of getting two numbers less than 4?
A. 1:3
B. 3:1
C. 1:4
D. 4:1
Answer:
c 1:4
Step-by-step explanation:
For 8-10, Find the area of the polygon.
Circle 1 is centered at (-4,
2) and has a radius of 3 centimeters. Circle 2 is centered at (5, 3) and has a radius of 6 centimeters.
What transformations can be applied to Circle 1 to prove that the circles are similar?
Enter your answers in the boxes.
The circles are similar because you can
translate Circle 1 using the transformation rule C, ) and then
dilate it using a scale factor of
The transformation for circle 1 exists (x+9), (y+5) and the scale factor of circle 1 exists 2.
What is a scale factor?Scale exists described as the ratio of the length of any object on a model to the actual length of the exact object in the entire world.
What is the transformation rule?The function transformation rules: f(x)+b changes the function b units upward. f(x)−b shifts the function b units downward. f(x + b) shifts the function b units to the left.
Circle 1
center: (−4, −2)
Radius: 3 centimeters
Circle 2
center: (5, 3)
Radius: 6 centimeters
Transformations can be used to circle 1 to verify that the circles exist similar.
As circle 2 and circle 1 do not contain the exact coordinates
So, circle 1 has to utilize transformation rules.
The difference between the coordinates of circle 2 and circle 1 exists
[tex]$x_2 - x_1 = 5 - (-4 ) = 9[/tex]
[tex]y_2 - y_1 = 3 - (-2) = 5[/tex]
Therefore, transformation for circle 1: (x+9), (y + 5)
The scale factor between circle 2 and circle 1 exists
The radius of Circle 2 = 2 [tex]*[/tex] radius of circle 1
Therefore, the scale factor of circle 1 = 2
Therefore, the transformation for circle 1 exists (x+9), (y+5), and the scale factor of circle 1 exists 2.
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Help please
please
Please
Gradient = rise/ run
rise is d
run is the difference in our x coordinates of c & e
(or c - e)
So, it's d/C - e
(3rd option)
Hope this helps!
Answer:
3rd option
Step-by-step explanation:
calculate the slope m using the slope formula
m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]
with (x₁, y₁ ) = E (e, 0 ) and (x₂, y₂ ) = (c, d )
m = [tex]\frac{d-0}{c-e}[/tex] = [tex]\frac{d}{c-e}[/tex]
Women athletes at a certain university have a long-term graduation rate of 67%. Over the past several years, a random sample of 36 women athletes at the school showed that 23 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from the university is now less than 67%
No this does not indicate that the population proportion of women athletes who graduate from the university is now less than 67%.
Given long term graduation rate of 67%, sample size of 36 and the women athletes graduated is 23.
We have to find whether the given information shows that the population proportion is less than 67%.
First we have to create hypothesis for this :
[tex]H_{0}[/tex]:P=0.67
[tex]H_{1}[/tex]:P<0.67
Under null hypothesis the test statistic is
z=p bar-p/[tex]\sqrt{p(1-p)/n}[/tex]
where p bar=23/36
=0.638
z=(0.638-0.67)/[tex]\sqrt{0.67(1-0.67)/36[/tex]
=-0.032/[tex]\sqrt{0.67*0.33/36}[/tex]
=-0.032/[tex]\sqrt{0.0064}[/tex]
=-0.032/0.078
=-0.41
Now we have to find the left tailed critical at 0.01 significance level using z table.
z=-2.33
Since the z value does not fall in the critical region,therefore we fail to reject the null hypothesis. So we can conclude that there is not sufficient evidence to say that the population proportion of women athletes who graduate from the university is now less than 67%.
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Question is incomplete as it should specify the signficance level of 0.01 to be used.
Olympia ate lunch at a restaurant the amount of her check was $6.89 she left eight dollars on the table which included the amount she owed plus a tip for the waiter which equation shows t and the amount of her tip in dollars
The equation will be like this,
6.89 + t = 8.00
t = 8 - 6.89
t = 1.11
tips = $1.11
6.89 + t = 8.00
To sum up the total amount of cash and coins you have, first, sort each invoice and coin by denomination. Create a separate stack for each denomination and count how many denominations or coins you have.
For each denomination of banknotes and coins, multiply the number you have by the denomination. For example, if you have 4 out of $ 10, multiply by 4x10 to get $ 40. If you have three $ 5 bills, multiply by 3x5 to get $ 15.
Sum the totals to calculate the total amount.
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Please help answer now please
Answer:
- 4/7 , -1/9, 7/4
Step-by-step explanation:
You know that 7/4 is the largest number because it is the only positive number. To compare the two negative numbers you need common denominators. The common denominator would be 133 (19 x 7). Now make equivalent fractions.-4/7 = x/133 133 divided by 7 is 19. Multiple the -4 by 19 to get -76/133. Next, make an equivalent fraction for -1/19. The new denominator will be 133. If I divide 133 by 19, I will get 7, so I multiple -1x7 and get -7 and the new fraction -7/133.
I have two fractions to compare -7/133 and -76/133. -76/133 if farthest from 0 so it has the least value.
is P - 1 upon p = 4 find P + 1 upon p whole square
Given that (p - 1/p) = 4, the value of p² + 1/p² is 18. Detail below
Data obtained from the questio(p - 1/p) = 4p² + 1/p² = ?How to determine the value of p² + 1/p²(p - 1/p) = 4
Square both sides
(p - 1/p)² = (4)²
(p - 1/p)² = 16 ....(1)
Recall
(a - b)² = a² + b² - 2ab
Thus,
(p - 1/p)² = p² + 1/p² - (2 × p × 1/p)
(p - 1/p)² = p² + 1/p² - 2
From equation (1) above,
(p - 1/p)² = 16
Therefore,
p² + 1/p² - 2 = 16
Rearrange
p² + 1/p² = 16 + 2
p² + 1/p² = 18
Thus, the value of p² + 1/p² is 18
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Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a)
Rolle's Theorem does not apply to the function because there are points on the interval (a,b) where f is not differentiable.
Given the function is [tex]f(x)=\sqrt{(2-x^{\frac{2}{3}})^{3}}[/tex] and the Rolle's Theorem does not apply to the function.
Rolle's theorem is used to determine if a function is continuous and also differentiable.
The condition for Rolle's theorem to be true as:
f(a)=f(b)f(x) must be continuous in [a,b].f(x) must be differentiable in (a,b).To apply the Rolle’s Theorem we need to have function that is differentiable on the given open interval.
If we look closely at the given function we can see that the first derivative of the given function is:
[tex]\begin{aligned}f(x)&=\sqrt{(2-x^{\frac{2}{3}})^3}\\ f(x)&=(2-x^{\frac{2}{3}})^{\frac{3}{2}}\\ f'(x)&=\frac{3}{2}(2-x^{\frac{2}{3}})^{\frac{1}{2}}\cdot \frac{2}{3}\cdot (-x)^{\frac{1}{3}}\\ f'(x)&=\frac{-\sqrt{2-x^{\frac{2}{3}}}}{\sqrt[3]{x}}\end[/tex]
From this point of view we can see that the given function is not defined for x=0.
Hence, all the assumptions are not satisfied we can reach a conclusion that we cannot apply the Rolle's Theorem.
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Please answer quickly
1a) 256/9
2a) 0.0048
3a) 0.64
4a) 2/5
5a) 4.6875
1b) 125/343
2b) 32/3125
3b) 0.003125
4b) 48.875
5b) 1.39
Whew! Hope this helps :)
Part E
Considering the probability you found in part D, which option makes more sense for Jake to choose?
Explain your reasoning.
The option which makes more sense for Jake to choose is option (1) He can accept the tie, and the game is over.
How to determine the best option to choose?The complete question is added as an attachment
From the complete question, the probability in part D is
Probability = 33.5%
The options to choose one from are:
Option 1: He can accept the tie, and the game is over.Option 2: He can make one more throw. Jake wins if he earns at least 30 points on the throw, but he loses if he earns less than 30 points.The calculated probability is less than 0.5.
This means that Jake is more likely to lose than win
So, the best option for Jake is option (A) walk away
Hence, the option which makes more sense for Jake to choose is option (1) He can accept the tie, and the game is over.
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The volume of an object that is recorded as 46cubic cm which was 15% from the actual volume
Two possibilities --- see below
Step-by-step explanation:
So it could be GREATER than 46
.85 x = 46 x = 54.11 cm^3
Or it could be smaller than 46
46= (1.15) x x = 40 cm^3
Rewrite each multiplication or division expression using a base and an exponent
4^5 divided by 4^2=
Using the laws of indices , 4⁵ ÷ 4² can rewritten as 4³.
What is the result of the division in exponent form?
Given the expression; 4⁵ ÷ 4²
To perform the division operation, we use one of the laws of indices;
[tex]n^a/n^b = n^{a-b}[/tex]
Given that;
n = 4a = 5b = 2Now, we apply the laws of indices ;
4⁵ ÷ 4² = 4⁵⁻² = 4³
Therefore, using the laws of indices, 4⁵ ÷ 4² can rewritten as 4³.
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Prove the following trigonometric identities
Answer:
Greetings !
check the attachment above☝️ but i haven't done the second question wait a moment. thx
Find the equation of the line which passes through the point ( 3,-2) and parallel to 2y - x = 3
The equation of the parallel line is 2y-x=-7
A parallel line will have the same slope, but different intercept. Let's assume C in the intercept of this parallel line. Therefore, equation of the parallel would be,
2y-x=C, coefficients of x and y remain the same, since parallel lines.
This line passes through (3, -2). Therefore, substituting those values, we will get value of C.
2.(-2) -3=C
C=(-7)
The equation of the parallel line is therefore 2y-x=-7
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Which expressions are equivalent to the given expression 5 log 10 x + log10 20 - log10 10
The expression which is equivalent to [tex]5 log_{10}x+log_{10}20-log_{10}10[/tex] is
[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-1[/tex].
What is the logarithmic equation?A logarithmic equation exists as an equation that applies the logarithm of an expression having a variable.
Product Rule Law: [tex]log_{a} (MN) = log_{a} M + log_{a} N[/tex]
Quotient Rule Law: [tex]log_{a} (M/N) = log_{a} M - log_{a} N[/tex]
Power Rule Law: [tex]log_{a}M^{n} = n log_{a} M[/tex]
Given: [tex]5 log_{10}x+log_{10}20-log_{10}10[/tex]
[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5) +log_{10}20 -log_{10}10[/tex]
apply the law of logarithm
[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5*20/10)[/tex]
[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5*2)[/tex]
[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(2x^5)[/tex]
Another possible equivalent expression is:
[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5*20)-log_{10}10[/tex]
[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-log_{10}10[/tex]
Substitute the value of [tex]log_{10}10 = 1[/tex]
[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-1[/tex]
Therefore, the correct answer
[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-1[/tex].
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