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4 years ago

In 2 + In 8 - In 4 A in 4 B in 6 C in64

Mathematics

2 answers:

kow [346]3 years ago

8 0

Answer:

Step-by-step explanation:

gayaneshka [121]3 years ago

8 0

C. in64

Explain.
Nintendo 64

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What is 3n+2n simplified

Ber [7]

The answer for this is n(3+2)

8 0

4 years ago

Read 2 more answers

The federal government requires a car manufacturer to have a minimum miles per gallon (mpg) average over the models it makes. Su

valentinak56 [21]

Answer:

25 mpg

Step-by-step explanation:

We have a set of values 19, 21, 22, 28 and 18

The average (which is the mean of the set of values ) is :

19 + 21 + 22 + 28 + 18 = 21.6 mpg out of the government requirement

So we need a value such as the mean become at least 25 mpg

We have to solve the equation:

(19+21+22+28+ x)/ 5 = 25

(90 + x ) /5 = 25 ⇒ 90 + x = 125 ⇒ x = 35 mpg

If we get the mean of the set of values including 35 we have:

(19 +21+ 22+ 28 + 35 ) / 5 = 25 mpg government requirement

7 0

3 years ago

What can you say about the end behavior of the function f(x)=-4^6+6x^2-52 ????

frosja888 [35]

Letter C is correct.
When the function is even, (all powers are even) both ends will go in the same direction.
In this case since the leading coefficient is negative, both ends will go down.

3 0

3 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$