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3 years ago

Find the lateral area of an equilateral triangular prism if the area of each lateral face is 10 square centimeters

1 answer:

3 0

The face dimensions are all the same, so the 3 lateral faces each have the same area. The lateral area is
3·(10 cm²) = 30 cm²

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Which statements are true about David's work? Check all that apply. The GCF of the coefficients is correct. The GCF of the varia

Anon25 [30]

Answer:

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

In step 6, David applied the distributive property.

Step-by-step explanation:

Given the polynomial :

80b⁴ – 32b²c³ + 48b⁴c

The Greatest Common Factor (GCF) of the coefficients:

80, 32, 48

Factors of :

80 : 1, 2, 4, 5, 8, 10, 16, 20, 40, and 80

32 : 1, 2, 4, 8, 16, and 32

48 : 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.

GCF = 16

b⁴, b², b⁴

b⁴ = b * b * b * b

b² = b * b

b⁴ = b * b * b * b

GCF = b*b = b²

GCF of c³ and c

c³ = c * c * c

c = c

GCF = c

We can see that David's GCF of the coefficients are all correct

From the polynomial ; 80b⁴ does not contain c ; so factoring out c is incorrect

In step 6 ; the distributive property was used to obtain ; 16b²c(5b² – 2c² + 3b²)

6 0

3 years ago

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What does y is two less than x stand for

solong [7]

Answer:

Y = X - 2 would be the answer.

5 0

3 years ago

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You call the pizza store and find out that each

qwelly [4]

The correct answer is 4

7 0

3 years ago

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I dont understand it :/

photoshop1234 [79]

Step-by-step explanation:

It's telling you to plug in the numbers inside the parenthesis as x

Example: F(-1)= 2(-1)^3 -3(-1)^2 +7

4 0

4 years ago

Triangle RST has vertices located at R (2, 3), S (4, 4), and T (5, 0).

LiRa [457]

The length of the sides are 2.24, 4.24 and 4.12 units while the slope are 0.5, -1 and -4.

<h3>Linear equation</h3>

A linear equation is in the form:

y = mx + b

where y, x are variables, m is the rate of change and b is the initial value of y.

For RS:

$Length = \sqrt{(y_2-y_1)^2+(x_2-x_1)^2}=\sqrt{(4-3)^2+(4-2)^2}=2.24 \\
\\
slope=\frac{y_2-y_1}{x_2-x_1} =\frac{4-3}{4-2}=0.5$

For RT:

$Length = \sqrt{(y_2-y_1)^2+(x_2-x_1)^2}=\sqrt{(0-3)^2+(5-2)^2}=4.24 \\
\\
slope=\frac{y_2-y_1}{x_2-x_1} =\frac{0-3}{5-2}=-1$

For ST:

$Length = \sqrt{(y_2-y_1)^2+(x_2-x_1)^2}=\sqrt{(0-4)^2+(5-4)^2}=4.12 \\
\\
slope=\frac{y_2-y_1}{x_2-x_1} =\frac{0-4}{5-4}=-4$

The length of the sides are 2.24, 4.24 and 4.12 units while the slope are 0.5, -1 and -4.

Find out more on linear equation at: brainly.com/question/14323743

5 0

2 years ago