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3 years ago

Which equation has a zero slope and y-intercept of -2?

Mathematics

1 answer:

slava [35]3 years ago

6 0

Answer:

It is D

Step-by-step explanation:

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4 pencils and 3 crayons are 20.9 inches. 5 pencils and 4 crayons are 26.5 inches. Each pencil is the same length as the other pe

erastova [34]

Answer:

The length of the pencils is 4.1 inches and for the crayons 1.5 inches

Step-by-step explanation:

We can solve by means of a 2x2 system of equations, we have to:

let "x" be the length of the pencils

let "y" be the length of the crayons

4 * x + 3 * y = 20.9

5 * x + 4 * y = 26.5 => x = (26.5 - 4 * y) / 5

Replacing:

4 * (26.5 - 4 * y) / 5 + 3 * y = 20.9

21.2 - 3.2 * y + 3 * y = 20.9

-0.2 * y = 20.9 - 21.2

y = -0.3 / -0.2

y = 1.5

now for x:

x = (26.5 - 4 * 1.5) / 5

x = 4.1

This means that the length of the pencils is 4.1 inches and for the crayons 1.5 inches

5 0

3 years ago

Assignment

Irina-Kira [14]

Manda una captura a ver

5 0

3 years ago

Read 2 more answers

Please Help!! (brainliest fr best answer)

lys-0071 [83]

Answer:

1. E=75 degrees

Step-by-step explanation:

7 0

3 years ago

Compare and contrast multiplication prop of equality and inequality

Alex

Answer:

Difference between multiplication property of equality and inequality

The only difference is: If you multiply or divide both sides of an equation by the same negative number, the equation remains the same, but If you multiply or divide both sides of an inequality by the same negative number, the inequality reverses. !!!!!

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

HELPP!!!!!!!!!!!!!!!!

Anvisha [2.4K]

Answer:

Step-by-step explanation:

We'll take this step by step. The equation is

$8-3\sqrt[5]{x^3}=-7$

Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time. First thing is to subtract 8 from both sides:

$-3\sqrt[5]{x^3}=-15$

The goal is to isolate the term with the x in it, so that means that the -3 has to go. Divide it away on both sides:

$\sqrt[5]{x^3}=5$

Let's rewrite that radical into exponential form:

$x^{\frac{3}{5}}=5$

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

$(x^{\frac{3}{5}})^{\frac{5}{3}}=5^{\frac{5}{3}}$

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely. On the right, let's rewrite that back in radical form to solve it easier:

$x=\sqrt[3]{5^5}$

Let's group that radicad into groups of 3's now to make the simplifying easier:

$x=\sqrt[3]{5^3*5^2}$ because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:

$x=5\sqrt[3]{5^2}$ which is the same as:

$x=5\sqrt[3]{25}$

8 0

3 years ago