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3 years ago

What is the value of the following expression? (−81)(−9)

Mathematics

2 answers:

neonofarm [45]3 years ago

3 0

Answer:

(−81)(−9) = 729

Step-by-step explanation:

Multiply -81 by -9 and you get 729

Kipish [7]3 years ago

3 0

Answer:

729

Step-by-step explanation:

The expression (−81)(−9) will give the value of 729. Step-by-step explanation: The given expression is (−81)(−9) .

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What is the solution to the system of equations?

andrew-mc [135]

D.(2,-3) if you substitute the numbers for the variables

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4 years ago

Find g(a - 1) when g(x) = 5x - 4.

lesya692 [45]

Answer:

${ \tt{g(x) = 5x - 4}}$

• solve when x is a - 1:

${ \tt{g(a - 1) = 5(a - 1) - 4}} \\ \\ { \tt{g(a - 1) = 5a - 5 - 4}} \\ \\ { \boxed{ \tt{g(a - 1) = 5a - 9 \: }}}$

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3 years ago

Find the dot product between the vectors u and v: u=2i+3j and v=-6i-5j

vitfil [10]

$\bf \stackrel{\textit{dot product}}{\cdot }\implies (a\cdot c)+(b\cdot d) \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} u=&2i+3j\\ v=&-6i-5j \end{cases}\implies \begin{cases} \\ \end{cases} \\\\\\ \cdot \implies (2\cdot -6)+(3\cdot -5)\implies -27$

4 0

3 years ago

What are the coordinates of the image of the point (1-, 2) under a dilation of 3 with the origin

Gemiola [76]

Answer:

The image of the point (1, -2) under a dilation of 3 is (3, -6).

Step-by-step explanation:

Correct statement is:

<em>What are the coordinates of the image of the point (1, -2) under a dilation of 3 with the origin.</em>

From Linear Algebra we get that dilation of a point with respect to another point is represented by:

$\vec P' = \vec R + r\cdot (\vec P-\vec R)$ (Eq. 1)

Where:

$\vec R$ - Reference point with respect to origin, dimensionless.

$\vec P$ - Original point with respect to origin, dimensionless.

$r$ - Dilation factor, dimensionless.

If we know that $\vec R = (0,0)$ , $\vec P = (1, -2)$ and $r = 3$ , then the coordinates of the image of the original point is:

$\vec P' = (0,0) +3\cdot [(1,-2)-(0,0)]$

$\vec P' = (0,0) + 3\cdot (1,-2)$

$\vec P' = (3,-6)$

The image of the point (1, -2) under a dilation of 3 is (3, -6).

4 0

3 years ago

If f(x) is differentiable for the closed interval [-4, 0] such that f(-4) = 5 and f(0) = 9, then there exists a value c, -4 <

WITCHER [35]

<span>Mean value theorem

For $f:[a,b] \rightarrow \mathbb{R}$ exist $c \in (a,b)$ such that

$f'(c)=\dfrac{f(b)-f(a)}{b-a}$

$f'(c)=\dfrac{f(-4)-f(0)}{-4-0}=\dfrac{5-9}{-4}=\dfrac{-4}{-4}=\boxed{1}$ </span>

3 0

3 years ago