All categories

2 years ago

Please Please Please help!

1 answer:

4 0

Take the first equation and set it to equal x by subtracting 5y from both sides and multiplying by -1. Looks like this...

-x + 5y = 1
-5y -5y
-x = -5y + 1
-1(-x = -5y + 1)
x = 5y - 1 ... substitute this into the second equation you get.....

2(5y - 1) + 4y = -4 answer is #2

You might be interested in

Help please how do I do this

salantis [7]

The transaction that is being used is a reflection

4 0

2 years ago

Read 2 more answers

100 POINTS PRE-CALCULUS

Sonja [21]

Answer:

not sure for the 1st one but pretty sure 2nd one is B

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Heather brought $7.14 to school. She then spent $1.08 to buy lunch. How much money does she have left?

Wewaii [24]

Answer:

6.06

Step-by-step explanation:

Simple subtraction.

5 0

3 years ago

Read 2 more answers

Can someone help me please

goldenfox [79]

Answer:

$1. \quad\dfrac{1}{k^{\frac{2}{3}}}\\\\2. \quad\sqrt[7]{x^5}\\\\3. \quad\dfrac{1}{\sqrt[5]{y^2}}$

Step-by-step explanation:

The applicable rule is ...

$x^{\frac{m}{n}}=\sqrt[n]{x^m}$

It works both ways, going from radicals to frational exponents and vice versa.

The particular power or root involved can be in either the numerator or the denominator. The transformation applies to the portion of the expression that is the power or root.

7 0

3 years ago

3(12-3y^2) Help solve this please

HACTEHA [7]

Step 1. <span>Distribute

</span> $3*12+3*-3 y^{2}$
<span>
Step 2. Simplify </span> $3*12$ to $36$

$36+3*-3 y^{2}$

Step 3. Simplify $3*-3 y^{2}$ to $-9 y^{2}$

$36-9 y^{2}$

Done! :)

5 0

3 years ago