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3 years ago

If f(x) = 2/3 x-6, what is the value of f(-9)?

Mathematics

1 answer:

Keith_Richards [23]3 years ago

4 0

Answer:

-12

Step-by-step explanation:

f(-9) = 2/3(-9) + (-6)

f(-9) = -6 + (-6)

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The measure of angle A to the nearest tenth of a degree is - 19.5

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3 years ago

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Please help! : What is the actual length of a window that is 1 1/4 inches long on a drawing with a scale of 1 inch= 6 feet? This

BabaBlast [244]

$\bf \begin{array}{ccllll}
model&actual\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
1&6\\
1\frac{1}{4}&x
\end{array}\implies \cfrac{1}{1\frac{1}{4}}=\cfrac{6}{x}
\\\\\\
1\frac{1}{4}\implies \cfrac{1\cdot 4+1}{4}\implies \cfrac{5}{4}\qquad thus
\\\\\\
$

$\bf \cfrac{1}{\frac{5}{4}}=\cfrac{6}{x}\implies \cfrac{\frac{1}{1}}{\frac{5}{4}}=\cfrac{6}{x}\implies \cfrac{1}{1}\cdot \cfrac{4}{5}=\cfrac{6}{x}\implies \cfrac{4}{5}=\cfrac{6}{x}
\\\\\\
x=\cfrac{5\cdot 6}{4}$

7 0

3 years ago

6 * 10^8 is how many times as large as 2 * 10^3

REY [17]

Answer:

300000 or 3 * 10^5

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Find the length of MN.<br> A) 7 <br> B) 25 <br> C) 32 <br> D) 39

MrRissso [65]

Step-by-step explanation:

Answer C)

Explanation:

MN + NP = MP

MP = 39

Therefore, 4(x + 5) + 2x + 1= MP

So, 4(x + 5) + 2x + 1= 39

(4x + 20) + 2x +1 = 39

So, (4x + 2x) + (20 + 1) = 39

6x + 21 = 39

Addition changes to Subtraction

So, 6x = 39 - 21

6x = 18

Multiplication changes to Division

x= 18/6

x=3

Therefore MN = 4(x + 5)

So, 4(3 + 5) = 4(8)

4(8) = 32

Therefore, MN = 32

Hope this helps!

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3 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$