A picture frame is made with 40 inches of materail, which expression should i use to frind the area

Factor x^3 -7x^2 -5x+35 by grouping. what is the resulting expression

Anni [7]

Answer:

(x-7) (x^2-5)

Step-by-step explanation:

x^3 -7x^2 -5x+35

Make 2 groups

x^3 -7x^2 -5x+35

Factor x^2 from the first group and -5 from the second group

x^2 (x-7) -5(x-7)

Now factor (x-7) out

(x-7) (x^2-5)

8 0

3 years ago

Will give brainliest

dmitriy555 [2]

Answer:

Step-by-step explanation:

Given, 

Radius of the cylinder = 3cm

Height of the cylinder = 3cm

Therefore, 

Lateral surface area of the cylinder

$= 2\pi rh$

$= 2\pi \times 3 \times 3 {cm}^{2}$

$= 2\pi \times 9 {cm}^{2}$

$= 18\pi {cm}^{2}$

Hence,

The required value in the green box will be 18 (Ans)

4 0

3 years ago

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The heights of a certain type of tree are approximately normally distributed with a mean height p = 5 ft and a standard

arsen [322]

Answer:

A tree with a height of 6.2 ft is 3 standard deviations above the mean

Step-by-step explanation:

⇒ $1^s^t$ statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)

an X value is found Z standard deviations from the mean mu if:

$\frac{X-\mu}{\sigma} = Z$

In this case we have: $\mu=5\ ft$ $\sigma=0.4\ ft$

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.4-5}{0.4}=1$

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

⇒ $2^n^d$ statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean. (FALSE)

For X =4.6 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1$

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

⇒ $3^r^d$ statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean (FALSE)

For X =5.8 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2$

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒ $4^t^h$ statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

For X =6.2\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3$

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.

6 0

3 years ago

Examine the following steps. Which do you think you might use to prove the identity Tangent (x) = StartFraction tangent (x) + ta

Over [174]

Answer:

The correct options are;

1) Write tan(x + y) as sin(x + y) over cos(x + y)

2) Use the sum identity for sine to rewrite the numerator

3) Use the sum identity for cosine to rewrite the denominator

4) Divide both the numerator and denominator by cos(x)·cos(y)

5) Simplify fractions by dividing out common factors or using the tangent quotient identity

Step-by-step explanation:

Given that the required identity is Tangent (x + y) = (tangent (x) + tangent (y))/(1 - tangent(x) × tangent (y)), we have;

tan(x + y) = sin(x + y)/(cos(x + y))

sin(x + y)/(cos(x + y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y))

(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y))

(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y)) = (tan(x) + tan(y))(1 - tan(x)·tan(y)

∴ tan(x + y) = (tan(x) + tan(y))(1 - tan(x)·tan(y)

6 0

3 years ago

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Solve for s. s 3 = 2

tresset_1 [31]

S= 5? that's at the top of my head

5 0

3 years ago

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