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3 years ago

What expression is equivalent to 5-3(6x+2)

Mathematics

1 answer:

alex41 [277]3 years ago

3 0

Answer: −18x-1

Step-by-step explanation: Hope this helps

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Factor 6x^4 - 5x^2+12x^2-10 by grouping. what is the resulting expression?

Alina [70]

12 goes with 6 and 5 with 10

(6x^4+12x^2)+(-5x^2-10)
(6x^2)(x^2+2)+(-5)(x^2+2)
undistribute
(x^2+2)(6x^2-5)

3 0

3 years ago

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Catherine has $400 in her checking account. She writes a check for $600. What is the balance of her account?

nevsk [136]

Answer:

Her new balance is $-200

$400 - $600 is $-200

8 0

3 years ago

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Helpppppppppppppppppppppppp

777dan777 [17]

I think it is 5 11/24

5 0

3 years ago

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The length of a string in yards is a function f(n) of the length n in inches. Write a function rule for this situation. f(n) = 1

8090 [49]

Given:

The length of a string in yards is a function f(n) of the length n in inches

To find:

The function rule for the given situation.

Solution:

We know that,

12 inches = 1 yard

1 inch = $\dfrac{1}{12}$ yard

n inches = $\dfrac{n}{12}$ yards

It is given that the length of a string in yards is a function f(n) of the length n in inches. So,

$f(n)=\dfrac{n}{12}$

Therefore, the required function for the given situation is $f(n)=\dfrac{n}{12}$ .

8 0

3 years ago

Can someone help me solve this differentiation/tangent problem?

nirvana33 [79]

A)

$\bf g'(x)=\stackrel{product~rule}{2x\cdot f(x)+x^2\cdot f'(x)}\quad 
\begin{cases}
x=5\\
f(5)=5\\
f'(5)=5
\end{cases}
\\\\\\
g'(5)=2(5)\cdot f(5)+(5)^2\cdot f'(5)\implies g'(5)=50+500\\\\\\ g'(5)=550\\\\
-------------------------------\\\\
g(5)=(5)^2f(5)\implies g(5)=125
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}
\begin{cases}
x=5\\
y=125\\
\stackrel{m}{g'(5)}=550
\end{cases}\implies y-125=550(x-5)
\\\\\\
y-125=550x-2750\implies y=550x+400$

b)

$\bf h'(x)=\stackrel{quotient~rule}{\cfrac{f'(x)(x-6)~~-~~f(x)\cdot 1}{(x-6)^2}}\quad 
\begin{cases}
x=5\\
f(5)=5\\
f'(5)=5
\end{cases}
\\\\\\
h'(5)=\cfrac{f'(5)(5-6)~~-~~f(5)\cdot 1}{(5-6)^2}
\\\\\\
h'(5)=\cfrac{5(-1)-5}{(-1)^2}\implies h'(5)=-10\\\\
-------------------------------\\\\$

$\bf h(5)=\cfrac{f(5)}{5-6}\implies h(5)=\cfrac{5}{-1}\implies h(5)=-5
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\quad 
\begin{cases}
x=5\\
y=-5\\
\stackrel{m}{-10}
\end{cases}\implies y-(-5)=-10(x-5)
\\\\\\
y+5=-10x+50\implies y=-10x+45$

3 0

4 years ago