Maths, would it be congruent or alternate??

Answer:

50 minutes

Step-by-step explanation:

3:20 pm - 4:50 = 90 minutes

90 = 2a + 15 + a

90-15= 75

75= 2a + a

75 = 3a

75/3 = 25

25 x 2 = 50 minutes

Step-by-step explanation:

Look, you're taking these kinds of questions too serious and that's why you believe they're hard, now pay close attention:

the areas are defined in cm² but the radius is defined in cm, so you need find the positive root of cm².

[tex] \sqrt{ \frac{5500 {cm}^{2} }{220 {cm}^{2} } } = \frac{r}{5cm} = = > \\ 5 = \frac{r}{5} = = = > r = 25[/tex]

and the question wants the base area:

[tex]s = \pi {r}^{2} = = = > s = 3.14 \times {(25cm)}^{2} = = > 1963.5 {cm}^{2} [/tex]

1+3+5+7+9+11+13+15

=4+12+20+28

=16+48

=64

For a proportional relationship, the constant is found dividing all values of y by each respective value of x.

A proportional relationship is a function in which the output variable is given by the input variable multiplied by a constant of proportionality, that is:

y = kx

In which k is the constant of proportionality.

The constant can be represented as follows:

[tex]k = \frac{y}{x}[/tex]

Hence the constant is found dividing all values of y by each respective value of x.

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[tex]P(R \: then \: Y) = \frac{3}{10} \times \frac{1}{10} = \frac{3}{100} [/tex]

[tex]note \: that = \\ order \: matters \: (no \: factorial) \\ replacement \: (equal \: sample \: space)[/tex]

Answer:

c. 3/100

Step-by-step explanation:

there are 10 marbles

First draw (there are 3 red marbles)

[tex]P=\frac{3}{10}[/tex]

Second draw (there is one yellow marble)

[tex]P=\frac{1}{10}[/tex]

probability of the event:

[tex]P=(\frac{3}{10} )(\frac{1}{10} )=\frac{3}{100}[/tex]

Hope this helps

Considering right-angled triangle XYW, an approximate ratio WY/WX is equal to: A. 0.34.

Considering right-angled triangle XYW, we would apply the law of cosine to approximate the ratio WY/WX. Mathematically, the law of cosine is given by:

cos(θ) = Adj/Hyp

Where:

Substituting the given parameters into the formula, we have;

cos(θ) = WY/WX

cos(70) = 3.4/10

0.34 = 0.34.

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Complete Question:

Right triangles 1-2 and 3 are given with all their angle measures and approximate side lengths.

Use one of the triangles to approximate the ratio WY/WX

A. 0.34

B. 0.94

C. 1.06

D. 2.76​

Answer:

See attached image

Step-by-step explanation:

The solution to the inequality is (–2.5, ∞)

The inequality is given as:

5.1(3 + 2.2x) > –14.25 – 6(1.7x + 4)

Open the brackets

15.3 + 11.22x > –14.25 – 10.2x - 24

Collect the like terms

11.22x + 10.2x > -15.3 - 14.25- 24

Evaluate the like terms

21.42x > -53.55

Divide both sides by 21.42

x > -2.5

This can also be represented as (–2.5, ∞)

Hence, the solution to the inequality is (–2.5, ∞)

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Answer:

c

Step-by-step explanation:

on edge

Answer:

54 degrees

Step-by-step explanation:

The actual angles of the triangle are not changing since this is just a reflection. The angles inside a triangle all add up to 180, so we can do 180-59-67 to get 54.

Let [tex]n=abcba[/tex] be such a number. If 18 divides [tex]n[/tex], then both 2 and 9 divide [tex]n[/tex].

To be divisible by 2, we must have [tex]a\in\{2,4,6,8\}[/tex]. Meanwhile we can have [tex](b,c)\in\{0,1,2,\ldots,9\}^2[/tex].

To be divisible by 9, we the sum of the digits of [tex]n[/tex] must itself be divisible by 9, or

[tex]2a+2b+c=9k[/tex]

for some integer [tex]k[/tex].

The largest value of [tex]2a+2b+c[/tex] is 2•8 + 2•9 + 9 = 43, so we must have [tex]k\in\{1,2,3,4\}[/tex].

I'm not sure what the best way to get the final count may be, but there are 44 such numbers. It's rather tedious to do by hand.

• If [tex]k=1[/tex], then [tex]2a+2b+c=9[/tex], and we can do this in 4 ways.

For example,

2•2 + 2•0 + 5 = 9 [tex](n = 20502)[/tex]

• If [tex]k=2[/tex], then [tex]2a+2b+c=18[/tex] and can be done in 16 ways.

2•2 + 2•3 + 8 = 18 [tex](n = 23832)[/tex]

• If [tex]k=3[/tex], then [tex]2a+2b+c=27[/tex] and can be done in 18 ways.

2•2 + 2•7 + 9 = 27 [tex](n = 27972)[/tex]

• If [tex]k=4[/tex], then [tex]2a+2b+c=36[/tex] and can be done in 6 ways.

2•6 + 2•8 + 8 = 36 [tex](n = 68886)[/tex]

The numbers : 4,-10 and -4 are integers.

A number without a decimal or fractional element is known as an integer, which can be both positive and negative, including zero.

The Latin word "integer" signifies "whole" or "intact." Thus, fractions and decimals are not included in integers.

All whole numbers and negative numbers are considered integers. This means that if we combine negative numbers with whole numbers, a collection of integers results.

"Negative 1 and one-third" includes a fractional part, so it is not an integer.

"2.5" and "0.ModifyingAbove 13 with Bar" contain a decimal, so it is also not an integer.

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Mathematical logic is important as it's a way to learn new experience through continuous self assessment.

Logic is important as it enables us to form sound judgements and beliefs.

The study of logic can help as it can help us to understand disagreement and ambiguity.

It also helps us in making a reasonable emotional life.

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It is determined while using the binomial distribution that there is still a 1.145=114.5% chance that she produces no more than 11 of them.

There are just two possible results for each throw. Either she succeeds or she fails. The binomial probability distribution is employed to answer this issue since the probability of completing a shot is regardless of all other throws.

Binomial probability distribution-

[tex]P(X=x) = C_{n,x} .p^{x}(1-p)^{n-x}[/tex]

[tex]C_{n,x} = n!/x! (n-x)![/tex]

where,

the no. of success= x

the no. of trials = n

the probability of a success on one trial = p

The probability of throwing not more than 11 will be:

P(X<11) = P(X=0) + P(X=1) +P(X=2) + P(X=3) +P(X=4) +P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)+P(X=11)

Where,

[tex]P(X=x) = C_{n,x} .p^{x}(1-p)^{n-x}[/tex]

[tex]P(X=0) = C_{18,0} .(0.689)^{0}(0.311)^{18}[/tex]≈0

[tex]P(X=1) = C_{18,1} .(0.689)^{1}(0.311)^{17}[/tex]≈0

[tex]P(X=2) = C_{18,2} .(0.689)^{2}(0.311)^{16}[/tex]≈0

[tex]P(X=3) = C_{18,3} .(0.689)^{3}(0.311)^{15}[/tex]≈0

[tex]P(X=4) = C_{18,4} .(0.689)^{4}(0.311)^{14}[/tex]≈0

[tex]P(X=5) = C_{18,5} .(0.689)^{5}(0.311)^{13}[/tex]= 0.0003

[tex]P(X=6) = C_{18,4} .(0.689)^{6}(0.311)^{12}[/tex]=0.0016

[tex]P(X=7) = C_{18,7} .(0.689)^{7}(0.311)^{11}[/tex]=0.0062

[tex]P(X=8) = C_{18,8} .(0.689)^{8}(0.311)^{10}[/tex]=0.0188

[tex]P(X=9) = C_{18,9} .(0.689)^{9}(0.311)^{9}[/tex]=0.0463

[tex]P(X=10) = C_{18,10} .(0.689)^{10}(0.311)^{8}[/tex]=0.9232

[tex]P(X=11) = C_{18,11} .(0.689)^{11}(0.311)^{7}[/tex]=0.1488

So,

P(X<11) = P(X=0) + P(X=1) +P(X=2) + P(X=3) +P(X=4) +P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)+P(X=11)

=0+0+0+0+0+0.0003+0.0016+0.0062+0.0188+0.0463+0.9232+0.1488 =1.145

Therefore, she makes 1.145=114.5% probability, no more than 11 of them.

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The correct option for the matrix will be:

a) If an n x n matrix A has fewer than n distinct eigenvalues, then A is not diagonalizable.

It could have repeated eigenvalues as long as the basis of each eigenspace is equal to the multiplicity of that eigenvalue.

b) If A is diagonalizable the A2 is diagonalizable

If A is diagonalizable then there exists an invertible matrix

c) If Rn has a basis of eigenvectors of A, then A is diagonalizable.

d) A is diagonalizable if and only if A has n eigenvalues, counting multiplicity.

e) If A is diagonalizable, then A is invertible.

It’s invertible if it doesn’t have a zero as eigenvalue but this doesn’t affect diagonalizable.

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Answer:

  7

Step-by-step explanation:

A 2-way table can be useful for recording the given information and for finding the missing numbers.

The attached table shows the numbers of families booking in July and August, and also counted by destination. Totals are at the right and bottom, and the grand total is the number 100 at the lower right.

Underlined numbers are those in the problem statement. The remaining numbers are computed so as to make the totals be correct.

Overall, 30+35 = 65 went to Portugal or France, so 100-65 = 35 went to Spain. Of those, 20 booked in August, so 15 booked to Spain in July.

59 booked in August, so a total of 41 booked in July.

Now we know 19 booked for France and 15 booked for Spain in July, so the remaining 7 booked for Portugal in July.

Answer:

8.048 liters

Step-by-step explanation:

Convert 8542 ml to liters:  (8542 ml)(1 liter/1000 ml) = 8.542 liters

Now subtract 0.458 liters to find the amount that a fish tank can hold:

 8.542 liters

-0.458  liters

8.048 liters is the fish tank capacity

The commission earned are R41123.17, R47931.67 and R15024.09

The VAT included prices are given as:

R271 800,00; R316 800,00 and R99 300,00.

The VAT rate is 12.36%.

So, the price before VAT is calculated as:

VAT included = Price without VAT * (1 + 12.36%)

This gives

VAT included = Price without VAT * 1.1236

So, we have:

Price without VAT = VAT included/1.1236

For the three cars, we have:

Price without VAT = R271 800.00/1.1236 = R241901

Price without VAT = R316800.00 /1.1236 = R281951

Price without VAT = R99300.00/1.1236 = R88377

Hence, the prices of the cars before VAT was added are R241901, R281951 and R88377

This is calculated as:

Commission = Price * 17%

So, we have:

Commission = R241901 * 17% = R41123.17

Commission = R281951 * 17% = R47931.67

Commission = R88377 * 17% = R15024.09

Hence, the commission earned are R41123.17, R47931.67 and R15024.09

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The portion of the mixture that is fruit to the nearest hundredth is 10.00 ounces

12 ounce juice:

= 70/100 × 12

= 0.7 × 12

= 8.4 ounces

Water = 30%

16 ounces;

= 10/100 × 16

= 0.1 × 16

= 1.6 ounces

Portion of mixture that is fruits = 8.4 ounces + 1.6 ounces

= 10.00 ounces

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Answer:

[tex]2 \tan (6x)+2 \sec (6x)+\text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

[tex]\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}[/tex]

If the terms are multiplied by constants, take them outside the integral:

[tex]\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x[/tex]

Multiply by the conjugate of 1 - sin(6x) :

[tex]\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x[/tex]

[tex]\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x[/tex]

[tex]\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:[/tex]

[tex]\implies \sin^2 (6x) + \cos^2 (6x)=1[/tex]

[tex]\implies \cos^2 (6x)=1- \sin^2 (6x)[/tex]

[tex]\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x[/tex]

Expand:

[tex]\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x[/tex]

[tex]\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:[/tex]

[tex]\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x[/tex]

[tex]\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}[/tex]

[tex]\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}[/tex]

Simplify:

[tex]\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}[/tex]

[tex]\implies 2 \tan (6x)+2 \sec (6x)+\text{C}[/tex]

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Substitute [tex]y=6x[/tex] and [tex]dy=6\,dx[/tex] to transform the integral to

[tex]\displaystyle \int \frac{12}{1-\sin(6x)} \, dx = 2 \int \frac{dy}{1 - \sin(y)}[/tex]

Now substitute [tex]t=\tan\left(\frac y2\right)[/tex] and [tex]dt=\frac12 \sec^2\left(\frac y2\right) \, dy[/tex] to transform this to

[tex]\displaystyle 2 \int \frac{dy}{1 - \sin(y)} = 2 \int \frac1{1-\frac{2t}{1+t^2}}\cdot\frac{2\,dt}{1+t^2} = 4 \int \frac{dt}{(t-1)^2}[/tex]

Finally, substitute [tex]s = t-1[/tex] and [tex]ds=dt[/tex] to get

[tex]\displaystyle 4 \int \frac{dt}{(t-1)^2} = 4 \int \frac{ds}{s^2} = -\dfrac4s + C[/tex]

Now recover the antiderivative in terms of [tex]x[/tex].

[tex]\displaystyle \int \frac{12}{1-\sin(6x)} \, dx = -\frac4s + C \\\\ ~~~~~~~~ = -\frac4{t-1} + C \\\\ ~~~~~~~~ = -\frac4{\tan\left(\frac y2\right) - 1} + C \\\\ ~~~~~~~~ = \boxed{-\frac4{\tan(3x) - 1} + C}[/tex]

The expression that is not equivalent to (1 - sin²(x)) tan(-x) is D. (cos²x - 1)(cot -x).

It should be noted that (cos²x - 1)(cot -x). is not equivalent to the given expression.

This is illustrated as:

(cos²x - 1)(cot -x)

= (-sin²x) × (-cot x)

= sin²x × cosx/sin x = sinxcosx

In conclusion, the correct option is D.

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The expression _______ is not equivalent to (1 − sin2(x)) tan(-x).

a. (1 - cos^2(x)) cot(-x)

b. (cos^2(x) - 1) cot(x)

c. (sin^(x) - 1) tan(x)

d. (cos^2(x) - 1) cot(-x)

Step-by-step explanation:

since there are z fishes

the number of angelfish = 1 × z

4

= 1 z

4

Based on the calculations, the correct value of the test statistic is equal to 3.2.

For samples A and B, the hypothesis is given by:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

Since both samples have a normal distribution, we would use a pooled z-test to determine the value of the test statistic:

[tex]z = \frac{\bar{x_1} - \bar{x_2} -(\mu_1 - \mu_2) }{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_1^2}{n_1}} }[/tex]

Substituting the given parameters into the formula, we have;

[tex]z = \frac{33 - 25 -(0) }{\sqrt{\frac{3^2}{4} + \frac{4^2}{4}} }\\\\z = \frac{8 }{\sqrt{\frac{9}{4} + \frac{16}{4}} }\\\\z = \frac{8 }{\sqrt{\frac{25}{4} } }\\\\z = \frac{8 }{\frac{5}{2} }}\\\\z = 8 \times \frac{2}{5}[/tex]

z = 3.2.

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18% of the milk mixture and how much of the cream mixture should he use.

A dairyman wants to combine milk with 5 percent butterfat and cream with 75 percent butterfat to create a 56-liter combination. Butterfat should make up 53% of the final combination.

Let x represent the volume of MILK required for the combination, in liters.

Consequently, x/56 is the PROPORTION of milk in the mixture. [because the final mixture has 56 liters in total]

We require 56 - x liters of CREAM in the mixture because we have 56 liters overall in the mixture.

The PROPORTION of cream in the combination is therefore equal to (56 - x)/56.

Our goal is for the final mixture to have 75% butterfat.

Fill in the equation with each of these values to obtain:

50 = (x/60)(5) + ((60 - x)/60) (75)

Add 56 to both sides to get: 3000 = (5)(x) + (56 - x)(75)

The formula is:

Multiply both sides by 56 to get: 3000 = (5)(x) + (56 - x)(75)

Expand: 3000 = 5x + 4200 - 75x

Simplify: 3000 = 4200 - 70x

Subtract 4500 from both sides: -1300 = -70x

Solve: x = (-1300)/(-70) = (1300)/(70) = 130/7

If you don't want to divide 130 by 7, you can evaluate this quickly by first realizing that 30/7 = 18.

Consequently, 130/7 must be a little larger than 18.

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Answer:

The width can either be 4 or 5 feet.

Step-by-step explanation:

Assuming the length is x and the width is y.2x+2y=18xy=202y=18−2xy=9−xx(9−x)=209x−x2=200=x2−9x+200=(x−5)(x−4)x=5and4 The width can either be 4 or 5 feet

The dosage at which the resulting blood pressure is maximized is x at 0.2 and the maximum dosage is 0.0008.

In the question,

The function is [tex]B(x) = 0. 06x^2 - 0. 2x^3[/tex]

To find the maximum or minimum, take the derivative and set it equal to zero.

⇒ [tex]B'(x) = 2(0. 06)x - 3(0. 2)x^{2}[/tex]

Setting it equal to zero, we get

⇒ [tex]0.12x - 0.6x^{2}=0[/tex]

⇒ 0.6x (0.2-x) = 0

⇒ x = 0 or x = 0.2

Now substitute x = 0.2 in B(x), we get

⇒ [tex]B(0.2) = 0. 06(0.2)^{2} - 0. 2(0.2)^{3}[/tex]

⇒ B(0.2) = 0.0024 - 0.0016

⇒ B(0.2) = 0.0008

To know B(0.2) is maximum, let us find the values for x = 1 and x = 0.01.

For x = 1,

⇒ [tex]B(1) = 0. 06(1)^{2} - 0. 2(1)^{3}[/tex]

⇒ B(1) = 0.06-0.2

⇒ B(1) = -1.04

For x = 0.01,

⇒ [tex]B(0.01) = 0. 06(0.01)^{2} - 0. 2(0.01)^{3}[/tex]

⇒ B(0.01) = 0.000006 - 0.0000002

⇒ B(0.01) = 0.0000058

Thus, x = 0.2 is the maximum.

Hence we can conclude that the dosage at which the resulting blood pressure is maximized is x at 0.2 and the maximum dosage is 0.0008.

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The discriminant of the quadratic equation is -20 and there are two non real solutions.

The discriminant of a quadratic equation uses the equation:

[tex]b^{2} -4ac[/tex]

Where the value of this calculation can tell you what solutions there are, plug known values in:

[tex]x^{2} +6x+14[/tex]

a=1

b=6

c=14

[tex]b^{2} -4ac[/tex]

which is equal to -20 on putting the values of a,b and c in the equation

As -20 < 0, this means that there is not a real solution, resulting in the first option being correct.

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The solution of the equation are as follows:

x = 6 and y = -5

4x + 5y = -1

-5x - 8y = 10

Therefore,

20x + 25y = -5

-20x - 32y = 40

-7y = 35

y = -5

Hence,

4x + 5(-5) = -1

4x - 25 = -1

4x = -1 + 25

4x = 24

x = 24 / 4

x = 6

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Answer:

Ralph Warren purchased 27 shares of stock at 16 3/8 per share. He paid a $27.50 brokerage fee. He later sold all 27 shares at 17 5/8 and paid a $28.75 brokerage fee. (36) What was his total cost for the stock including his brokerage fee? (37) What did he receive from the sale of the stock after he paid the brokerage fee? (38) Did he have a capital gain or loss? (39) How much was the gain or loss? (40) What was the net change from 16 3/8 to 17 5/8?

Step-by-step explanation: