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rodikova [14]
3 years ago
14

What is the answer of this equation ? 10x-6(2x+5)=20

Mathematics
1 answer:
Rudiy273 years ago
6 0

Answer:

Simplifying

x = -25

Step-by-step explanation:

Reorder the terms:

10x + -6(5 + 2x) = 20

10x + (5 * -6 + 2x * -6) = 20

10x + (-30 + -12x) = 20

Reorder the terms:

-30 + 10x + -12x = 20

Combine like terms: 10x + -12x = -2x

-30 + -2x = 20

Solving

-30 + -2x = 20

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '30' to each side of the equation.

-30 + 30 + -2x = 20 + 30

Combine like terms: -30 + 30 = 0

0 + -2x = 20 + 30

-2x = 20 + 30

Combine like terms: 20 + 30 = 50

-2x = 50

Divide each side by '-2'.

x = -25

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Nonamiya [84]

Use the substitution method

Answers:

2m

2(1)

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2(2)

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7 0
3 years ago
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padilas [110]
The answer is edition
8 0
3 years ago
Find f such that the given conditions are satisfiedf’(x)=x-4, f(2)=-1
kicyunya [14]

Given:

f^{\prime}\left(x\right)=x-4,\text{ and}f\left(2\right)=-1

To find:

The correct function.

Explanation:

Let us consider the function given in option D.

f(x)=\frac{x^2}{2}-4x+5

Differentiating with respect to x we get,

\begin{gathered} f^{\prime}(x)=\frac{2x}{2}-4 \\ f^{\prime}(x)=x-4 \end{gathered}

Substituting x = 2 in the function f(x), we get

\begin{gathered} f(2)=\frac{2^2}{2}-4(2)+5 \\ =2-8+5 \\ =-6+5 \\ f(2)=-1 \end{gathered}

Therefore, the given conditions are satisfied.

So, the function is,

f(x)=\frac{x^{2}}{2}-4x+5

Final answer: Option D

6 0
1 year ago
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Mekhanik [1.2K]

Answer:

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Step-by-step explanation:

Because there are no red marbles

6 0
3 years ago
Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)
Leni [432]
Alrighty

remember some rules
(ab)^c=(a^c)(b^c)
and
x^{-m}=\frac{1}{x^m}
and
x^0=1 for all real values of x
and
(a^b)^c=a^{bc}
and
(\frac{a}{b})^c=\frac{a^c}{b^c}
and
(a/b)/(c/d)=(ad)/(bc)
and
(a^b)(a^c)=a^{b+c}
and
\frac{a^m}{a^n}=a^{m-n}
and don't forget pemdas
example: -x^m=-1(x^m) but (-x)^m=(-1)^m(x^m)

so

4.
(\frac{c^{-2}}{2})^{-2}=

\frac{(c^{-2})^{-2}}{2^{-2}}=

\frac{c^4}{\frac{1}{2^2}}=

\frac{c^4}{\frac{1}{4}}=

4c^4



5.

\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=

\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=

\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=

\frac{(a^4bc^5)b^3}{(-1)(a^2)}=

\frac{-a^4b^4c^5}{a^2}=

-a^2b^4c^5
3 0
3 years ago
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