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2 years ago

What does 15x+35 equal

Mathematics

1 answer:

____ [38]2 years ago

7 0

15x+35 = 5(3x + 7)

hope that helps

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Choose the congruence theorem that you would use to prove the triangles congruent.

Evgesh-ka [11]

The answer is SSS . This is because the sides values are shown.

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3 years ago

How many are 16 x 16 ?

yan [13]

Answer:

256

Step-by-step explanation:

16 ×16= 256

Hope it works out for you

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3 years ago

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Find the complete factorization of the expression. 16xz + 20yz

Andreyy89

To completely factor you must from the Greatest common factor

In this case it is 4z

So the complete factorization is

4z(4x + 5y)

8 0

3 years ago

I will give 30 points if u help now

Amiraneli [1.4K]

Answer:

3. x=16

4. x=51

Step-by-step explanation:

3. 3x-8=x+24

2x=32

x=16

4. 2/3(x+27)= 180-(3x-25)

2/3x+18=180-3x+25

11/3x=187

11x=561

x=51

5 0

2 years ago

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Which value is NOT a solution of 8x3 – 1 = 0?

Tpy6a [65]

Note: As you may have unintentionally missed to add the value choices. But, I would make sure to explain the concept so that you may improve your understanding in terms of solving these type of questions.

Answer:

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Step-by-step explanation:

Considering the equation

$8x^3\:-\:1\:=\:0$

Steps to solve the equation

$8x^3-1=0$

$\mathrm{Add\:}1\mathrm{\:to\:both\:sides}$

$8x^3-1+1=0+1$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

$\mathrm{Divide\:both\:sides\:by\:}8$

$\frac{8x^3}{8}=\frac{1}{8}$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

$\mathrm{For\:}x^3=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

$x=\sqrt[3]{\frac{1}{8}},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1-\sqrt{3}i}{2}$

So,

$x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$

Therefore,

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Keywords: solution, value

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7 0

2 years ago