All categories

4 years ago

How many solutions does 25-x=15-(3x+10)

1 answer:

5 0

25-x=15-(3x+10)
25-x=15-3x-10
25-x=15-10-3x
25-x=5-3x
25-5-x=5-5-3x
20-x=-3x
20-x+x=-3x+x
20=-2x
20/-2=-2/-2x
-10=x

Only one solution for this equation.

You might be interested in

X(y - 4)<br> I’m very bad at this

chubhunter [2.5K]

Answer: That would be xy−4x <3 hope it helps

Step-by-step explanation: all u have to do is multiply the polynomials

8 0

3 years ago

I need help with this problem.

emmasim [6.3K]

What nanajdisn at x Andorrans

8 0

3 years ago

Read 2 more answers

Solve the right triangle. round your answers to the nearest tenth. help.

hammer [34]

Answer:

19.2

Step-by-step explanation:

Cosine rule :

$\sqrt{a^{2} +b^{2} ﹣2abcos(angle ) }$

c will always be the hypotenuse.

a and b can be any if the opposite or the adjacent.

${15}^{2} + 12^{2} - 2 \times 15 \times 12 \times \cos(90) = 369$

\sqrt{369}=19.20937...

8 0

3 years ago

Alexis runs 8/10 miles a day. How many complete miles will Alexis finish in a week?

Assoli18 [71]

Answer:

5.6 Miles I believe

5 0

4 years ago

Can I please get some help with this

Marizza181 [45]

After solving $\sqrt[5]{128x^8y^2}$ we get $2x\sqrt[5]{4}\sqrt[5]{x^3}\sqrt[5]{y^2}$

Step-by-step explanation:

We need to solve $\sqrt[5]{128x^8y^2}$

Applying the rule: $\sqrt[a]{xy}=\sqrt[a]{x}.\sqrt[a]{y}$

$\sqrt[5]{128x^8y^2}\\=\sqrt[5]{128}\sqrt[5]{x^8}\sqrt[5]{y^2}\\We\,\,know\,\,that\,\,128=2\times2\times2\times2\times2\times2\times2=2^7\,\,or\,\,2^5.2^2\\=\sqrt[5]{2^5.2^2}\sqrt[5]{x^5.x^3}\sqrt[5]{y^2}\\=(2^5)^{\frac{1}{5}}\sqrt[5]{2^2}\sqrt[5]{x^5}\sqrt[5]{x^3}\sqrt[5]{y^2}\\=2x\sqrt[5]{4}\sqrt[5]{x^3}\sqrt[5]{y^2}$

So, After solving $\sqrt[5]{128x^8y^2}$ we get $2x\sqrt[5]{4}\sqrt[5]{x^3}\sqrt[5]{y^2}$

Keywords: Solving with Exponents

Learn more about Solving with Exponents at:

#learnwithBrainly

3 0

3 years ago