URGENT WHOEVER ANSWERS CORRECTLY AND FAST GETS BRAINLY
40% of the house plants were watered today. 320 house plants were watered. Which equation can be used to find how many plants there were?

40x = 320

0.40x = 320


x
=
0.40
--------
320

x = 0.40(320)

The function for the given graph is f(x) = (x - 4)(x + 2)(x + 4).

In the given graph,

The points that lie on the x-axis are (-4, 0), (-2, 0), and (4, 0) where y-coordinate is 0 and the x coordinates are -4, -2, and 4.

On solving the given functions,

Option A:

f(x) = (x - 4)(x + 2)(x + 4)

Since y = 0 consider f(x) becomes 0

So,

(x - 4)(x + 2)(x + 4) = 0

According to the zero product rule,

x - 4 = 0, x + 2 = 0, and x + 4 = 0

⇒ x = 4, x = -2 and x = -4.

Thus, option A represents the function of the given graph.

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Question:

Select the correct answer. which equation could represent the graphed function (given in the figure below)?

The function equality that represents the point where f(x) = g(x) is; A: f(0) = g(0) and f(2) = g(2)

From the question, the curved red line represents g(x) while the straight blue line represents f(x).

Now, the equality of functions f(x) = g(x)  is represented as a common function between their curves. Thus, we will just need to find a common point for both. This is;

f(x) has points (0, 4) and (2, 0).

g(x) has points (0,4) and (2,0).

It is pertinent to note that both functions have the same y-value for x=0 and x = 2. Thus, we can say that;

f(2) = g(2) and f(0) = g(0)

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Answer:

88

Step-by-step explanation: make me brainliest if it is correct

Answer:

[tex]\frac{FH}{RX} =\frac{HG}{XS} =\frac{FG}{RS}[/tex]

Step-by-step explanation:

The order of the letters matters in any similarity theorem.

When triangle FHG is similar to triangle RXS, the lengths of FH must be similar to RX.

When 2 similar lines are divided, that number is the factor by which one triangle can be dilated to match the other.

That being said, each of these proportion statements can be read as individual division expressions that equal the next division expression.

This means [tex]FH\div RX = HG\div XS = FG\div RS[/tex].

If you look at it this way, you can see that the sides you need to fill in must be the corresponding similar sides, so that you end up with the factor for the dilation of the 2 triangles.

Answer:

1

Step-by-step explanation:

This is the only problem that could be answered with this equation, because you would add the bracelets he made (2+4) then times that number by the days he made them, 5, then divide by the number of groups, 6. Which equals 5 x ( 2 + 4 ) divided by 6.

I hope this helps!

The inequality of the graph that is given is: y < 2x - 3.

First find the slope (m) = rise / run of the line.

Rise = 2 units

Run = 1 unit

Slope (m) = 2/1 = 2

Y-intercept (b) = -3.

Since the shaded side is at the left, we are going to use the ">" sign. Substitute m = 2 and b = -3 into y < mx + b.

y < 2x - 3

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Answer:

Option 1

Step-by-step explanation:

Find 2 numbers that multiply to give 16 and sum to give +8 :

These numbers are 4 and 4

Now split +8x into +4x and +4x and factor each part :

x²+4x+4x+16 = 0

x(x+4)+4(x+4) = 0

Keep 1 bracket and the terms outside the brackets make their own :

(x+4)(x+4) = 0

(x+4)²  = 0

Therefore answer will be Option 1

Hope this helped and have a good day

Answer:

24 combinations

Step-by-step explanation:

To get to this answer, I multiplied 6 x 4, since there are 6 shirts and 4 pairs of jeans, there will be a total of 24 combinations. For combinations of two different things, you can just multiply the amount of each item by the other item. hope this helps :)

Question 2

1) [tex]\overline{PR} \perp\overline{QS}, \overline{PQ} \cong \overline{PS}[/tex] (given)

2) [tex]\overline{PR} \cong \overline{PR}[/tex] (reflexive property)

3) [tex]\angle PRQ, \angle PRS[/tex] are right angles (perpendicular lines form right angles)

4) [tex]\triangle PRQ, \triangle PRS[/tex] are right triangles (a triangle with a right angle is a right triangle)

5) [tex]\triangle PRQ \cong \triangle PRS[/tex] (HL)

Question 3

1) [tex]\angle C[/tex] is a right angle, [tex]\overline{AC} \cong \overline{AE}, \overline{DE} \perp \overline{AB}[/tex] (given)

2) [tex]\angle DEA[/tex] is a right angle (perpendicular lines form right angles)

3) [tex]\triangle ACD, \triangle DAE[/tex] are right triangles (a triangle with a right triangle is a right angle)

4) [tex]\overline{AD} \cong \overline{AD}[/tex] (reflexive property)

5) [tex]\triangle ACD \cong \triangle AED[/tex] (HL)

6) [tex]\angle CAD \cong \angle DAE[/tex] (CPCTC)

7) [tex]\overline{AD}[/tex] bisects [tex]\angle BAC[/tex] (if a segment splits an angle into two congruent parts, it is an angle bisector)

Based on the length and width of the rectangular lawn, and the percentage the man has mowed, the strip width would be 40.5 m.

first, find the area of the lawn:

= 100 x 200

= 20,000 ft²

the area mowed is:

= 40.5% x 20,000

= 8,100 ft²

the width is therefore:

= 8,100 / 200

= 40.5 m

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The probability of randomly getting first a red marble and then a blue marble is:

P = 0.083

There are:

For a total of 9 marbles.

The probability of getting a red marble is equal to the quotient between the number of red marbles and the total, so:

P(red) = 2/9

Then the probability of getting a blue marble is equal to the quotient between the number of blue marbles and the total, but because we already took one marble, now the total is 8.

P(blue) = 3/8

The joint probability is given by the product:

P = (2/9)*(3/8) = 0.083

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Answer:

0.42

Step-by-step explanation:

The value is 52. 674

We have;

4(3) / 4(8) +.9(7.8 x 7.45)

From BODMAS, we have that we must must open up the bracket

= [tex]\frac{12}{32} + 0. 9 (58. 11)[/tex]

Divide the fraction to get decimals

= [tex]0. 375 + 52. 299[/tex]

Add the decimals

= [tex]52. 674[/tex]

Thus, the value is 52. 674

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Answer:

6y^2 + 18y

Step-by-step explanation:

2y (3y + 9)=

6y^2 + 18y

[2y (3y becomes 6y^2]

[2x(      + 9) becomes 18y]

Add them to obtain 6y^2 + 18y

The simplified form is [tex]6x^{3}[/tex].

What are indices in maths?

The equation is -

[tex]\sqrt{\frac{2160x^{3} }{60x^{2} } }[/tex]

We can rewrite this using property of multiplication to obtain

[tex]\sqrt{\frac{2160}{60} * \frac{x^{8} }{x^{2} } }[/tex]

We simplify variable expression in the radicand using

[tex]\frac{a^{m} }{a^{n} } = a^{m - n}[/tex]

[tex]\sqrt{36 * x^{8 - 2} }[/tex]

[tex]\sqrt{36 * x^{6} }[/tex]

[tex]\sqrt{36 * (x^{3})^{2} }[/tex]

[tex]\sqrt{36} * \sqrt{(x^{3})^{2} }[/tex]

[tex]6 * x^{3}[/tex]

Therefore,  the simplified form is [tex]6x^{3}[/tex]

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The lowest score you can earn and still be eligible for employment is 85.6925.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean  and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

Top 5%.

At least the 100-5 = 95th percentile.

The 95th percentile is X when Z has a pvalue of 0.95. So X when Z = 1.645.

x-75 = 1.645*6.5

x = 85.692

Thus the lowest score you can earn and still be eligible for employment is 85.6925.

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Given the 4th order linear ODE

[tex]u^{(4)} - u = 0[/tex]

we substitute

[tex]x_1 = u[/tex]

[tex]x_2 = {x_1}' = u'[/tex]

[tex]x_3 = {x_2}' = {x_1}'' = u''[/tex]

[tex]x_4 = {x_3}' = {x_2}'' = {x_1}''' = u'''[/tex]

Then the given equation transforms to

[tex]{x_4}' - x_1 = 0[/tex]

but we also need to relate this to the other derivative substitutions. This gives the system of differential equations

[tex]\begin{cases} {x_1}' = x_2 \\ {x_2}' = x_3 \\ {x_3}' = x_4 \\ {x_4}' = x_1 \end{cases}[/tex]

In matrix form,

[tex]\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}' = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}[/tex]

The system of linear equations can be used to find the number of nickels, n, and the number of quarters, q, yolanda used is n + q = 28

n + q = 280.05n + 0.25q = 4

n + q = 28

0.05n + 0.25q = 4

n = 28 - q

0.05n + 0.25q = 4

0.05(28 - q) + 0.25q = 4

1.4 - 0.05q + 0.25q = 4

- 0.05q + 0.25q = 4 - 1.4

0.20q = 2.6

q = 2.6 / 0.20

q = 13

Substitute into

n + q = 28

n + 13 = 28

n = 28 - 13

n = 15

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196 36 correct answer is 232

Answer:

728 cm²

Step-by-step explanation:

the surface area is the area of the 6 faces.

the opposite rectangular faces are congruent , then

SA = front/ back + sides + top/ bottom

     = 2(14 × 14) + 2(6 × 14) + 2(14 × 6)

     = 2 × 196 + 2 × 84 + 2 × 84

     = 392 + 168 + 168

     = 728 cm²

There are 24 ways in which 5 guys can sit if arranged from oldest to youngest.

We have,

Five guys.

Now,

We know that,

Total number of ways to arranged around a table (n) = (n-1)!

So,

For n = 5,

I.e.

(n - 1)! = (5 - 1)! = 4!

So,

Total number of ways to arranged Five guys rom oldest to youngest (n) = (n-1)!

i.e.

= (5 - 1)! = 4!

We get,

= 4 × 3 × 2 × 1

i.e.

Total number of ways to arranged Five guys rom oldest to youngest (n) = 24.

Hence we can say that there are 24 ways in which 5 guys can sit if arranged from oldest to youngest.

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Answer:

width = 9 inches

Step-by-step explanation:

let width be w then length is w + 12

the perimeter ( P) is calculated as

P = 2( width + length )

here P = 60 , then

2(w + w + 12) = 60 ( divide both sides by 2 )

2w + 12 = 30 ( subtract 12 from both sides )

2w = 18 ( divide both sides by 2 )

w = 9

that is the width of the rectangle is 9 inches

Answer:

Step-by-step explanation:

[tex]\sf 1.1)\dfrac{Sin \ \theta-Cos \ \theta}{Sin \ \theta + Cos \ \theta}=\dfrac{(Sin \ \theta-Cos \ \theta)}{((Sin \ \theta + Cos \ \theta)}*\dfrac{(Sin \ \theta-Cos \ \theta)}{(Sin \ \theta-Cos \ \theta)}[/tex]

                           [tex]\sf = \dfrac{(Sin \ \theta-Cos \ \theta)^2}{Sin^2 \ \theta-Cos^2 \ \theta}\\\\=\dfrac{Sin^2 \ \theta+Cos^2 \ \theta-2Sin \ \theta \ Cos\ \theta}{(Sin \ \theta-Cos \ \theta)}\\\\\\\bf Identity: \ (a +b)^2= a^2 + b^2 - 2ab\\\\\\=\dfrac{1-2Sin \ \theta \ Cos \ \theta}{(Sin \ \theta-Cos \ \theta)} = LHS[/tex]

[tex]\sf 1.2) LHS = tan^2 \ x - Sin^2 \ x = \dfrac{Sin^2 \ x}{Cos^2 \ x}-Sin^2 \ x[/tex]

                                        [tex]\sf =\dfrac{Sin^2 \ x}{Cos^2 \ x}-\dfrac{Sin^2 \ x*Cos^2 \ x}{1*Cos^2 \ x}\\\\\\ = \dfrac{Sin^2 \ x - Sin^2 \ x*Cos^2 \ x}{Cos^2 \ x}\\\\\\= \dfrac{Sin^2 \ x *(1 -Cos^2 \ x)}{Cos^2 \ x}\\\\=\dfrac{Sin^2 \ x*Sin^2 \ x}{Cos^2 \ x} \\\\ \bf 1 - Cos^2 \ x = Sin^2 \ x\\\\= \dfrac{Sin^2 \ x}{Cos^2 \ x}*Sin^2 \ x\\\\=tan^2 \ x * Sin^2 \ x = RHS[/tex]

[tex]\sf 1.3) LHS = \dfrac{1-Cos \ x}{Sin \ x}-\dfrac{Sin \ x}{1+Cos \ x} =\dfrac{(1-Cos \ x)(1+Cos \ x)}{Sin \ x*(1+Cos \ x)}-\dfrac{Sin \ x*Sin \ x}{(1+Cos \ x)*Sin \ x}\\[/tex]

              [tex]\sf =\dfrac{1 - Cos^2 \ x}{Sin \ x*(1+Cos \ x)}-\dfrac{Sin^2 \ x}{Sin \ x*(1+Cos \ x)}\\\\=\dfrac{Sin^2 \ x}{Sin \ x*(1+Cos \ x)} - \dfrac{Sin^2 \ x}{Sin \ x*(1+Cos \ x)}\\\\=\dfrac{Sin^2 x - Sin^2 \ x}{Sin \ x*(1+Cos \ x)} \\\\= 0 = RHS[/tex]

[tex]\sf 1.4) LHS = Sin x - \dfrac{1}{Sin \ x + Cos \ x}+Cos \ x \\[/tex]

              [tex]\sf = \dfrac{Sin \ x *(Sin \ x + Cos \ x) - 1 + Cos \ x * (Sin \ x + Cos \ x)}{Sin \ x + Cos \ x }\\\\\\= \dfrac{Sin \ x * Sin \ x + Sin \ x*Cos \ x -1 + Cos \ x*Sin \ x + Cos \ x*Cos \ x}{Sin \ x + Cos \ x}\\\\\\=\dfrac{Sin^2 \x + Sin \ x \ Cos \ x - 1 + Cos \ x \ Sin \ x + Cos^2 \x}{Sin \ x + Cos \ x}\\\\=\dfrac{Sin^2 \ x + Cos^2 \ x - 1 + Sin \ xCos \x +Sin \ x Cos \ x}{Sin \ x + Cos \ x}\\\\= \dfrac{1 - 1 +2Sin \ x Cos \ x}{Sin \ x + Cos \ x}\\\\= \dfrac{2Sin \ x Cos \ x}{Sin \ x + Cos \ x}[/tex]

The sum of areas of circle C and circle D exists 98 square units.

Given: Radius of circle = 7 units

Area of circle [tex]=\pi r^2[/tex] ..............(1)

Where, r be the radius of circle

The radius of circle C, r = 7 units

(1) becomes

Area of circle, C [tex]= \pi 7^2[/tex] square units

[tex]= 49\pi[/tex] units²

Area of circle, D [tex]= \pi 7^2[/tex] square units

[tex]= 49\pi[/tex] units²

Sum of areas of circles C and D,

= Area of circle C + Area of circle D

= 49 + 49

= 98 units²

The sum of areas of circles C and D exist 98 units².

Therefore, the correct answer is option d) 98 units².

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The correct options about parabola are:

The focus is located at (0,–3).

The parabola can be represented by the equation x^2 = –12y.

According to the statement

we have given that the vertex is at the origin and directrix at y = 3.

and from these given information and we have to find the all abot the parabola like its focus point etc.

So,

We know that the equation of parabola is

(x-h)^2 = 4p(y-k)

Here The vertex is (h,k). and the focus is at (h,k+p). and the directrix is   y(k - p.)

So, From the given information

Vertex at the origin means that h=0 and k=0

Directrix at y = 3 means that p=-3

Directrix at the y-axis means the parabola opens upwards.

Thus, the focus is: (0,-3)

And The p-value becomes

: 4(-3) = -12.

And from all these the equation of the parabola is becomes

(x)^2 = -12y

So, The correct options about parabola are:

The focus is located at (0,–3).

The parabola can be represented by the equation x^2 = –12y.

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Question:

A parabola, with its vertex at the origin, has a directrix at y = 3. Which statements about the parabola are true? Select two options.

The focus is located at (0,–3).

The parabola opens to the left.

The p value can be determined by computing 4(3).

The parabola can be represented by the equation x2 = –12y.

The parabola can be represented by the equation y2 = 12x.

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The solution to the expression [tex]8x^3-4x+\frac{2}{3x}-\frac{1}{27x^3}[/tex] is [tex]\frac{216x^6 - 108x^4 + 18x^2 - 1}{27x^3}[/tex]

The expression is given as:

[tex]8x^3-4x+\frac{2}{3x}-\frac{1}{27x^3}[/tex]

Take the LCM of the expression

[tex]\frac{8x^3 * 27x^3 - 4x * 27x^3 + 2 * 9x^2 - 1}{27x^3}[/tex]

Evaluate the products

[tex]\frac{216x^6 - 108x^4 + 18x^2 - 1}{27x^3}[/tex]

Hence, the solution to the expression [tex]8x^3-4x+\frac{2}{3x}-\frac{1}{27x^3}[/tex] is [tex]\frac{216x^6 - 108x^4 + 18x^2 - 1}{27x^3}[/tex]

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Since beta is in the first quadrant, the final answer will be positive.

To find cos(beta) so we can use the half angle identity, we can substitute into the Pythagorean identity. Doing so gives us that

[tex] \sin( \beta ) = \frac{ \sqrt{17} }{5} [/tex]

So, this means that

[tex] \sin( \frac{ \beta }{2} ) = \sqrt{ \frac{1 - \frac{ \sqrt{17} }{2} }{2} } = \sqrt{ \frac{2 - \sqrt{17} }{4} } = \frac{ \sqrt{2 - \sqrt{17} } }{2} [/tex]

Answer:

Step-by-step explanation:

Area = h/2 (b_1+b_2)

[tex]\sf\cfrac{7}{2}\:(22+34)[/tex]

Add 22 and 34 = 56

[tex]\sf \cfrac{7}{2} \times 56[/tex]

Let's express 7/2 * 56 as a single fraction:-

[tex]\sf \cfrac{7\times 56}{2}[/tex]

Multiply 7 and 56 = 392

[tex]\sf \cfrac{392}{2}[/tex]

Divide

[tex]\sf 196[/tex]

Therefore, the area of the trapezoid is 196cm².

_______________________

No, this not plagiarism .You can use generally accepted facts without citing them.

What defines self-plagiarism?

Self-plagiarism is defined as a type of plagiarism in which the writer republishes a work in its entirety or reuses portions of a previously written text while authoring a new work.

What is considered general knowledge in plagiarism?

How to Avoid plagiarism common knowledge?

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Answer:

Step-by-step explanation: