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3 years ago

G(-4)=? =>Khan Academy Algebra question<=

Mathematics

1 answer:

larisa [96]3 years ago

3 0

Maybe it means where point g is move it down 4 i could be wrong

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Can I get some help? <br><br> Please show your work! <br><br> Thanks!

kirill [66]

Hey there,

Your question states: You have a 332 feet fencing to enclose a rectangular region. What is the maximum area.

Your correct answer would be 6,889.

Reason/explanation:

The formula that we would use is $P = 2l+2w$

$f(x)=ax^{2} +b*x+c$

$L=166=-W\\ L=83$
_____________________________________________________________

So by this, now that we know that $L=83$

We do . . . 83 × 83. . .

And you final answer would be Option A.) 6,889

Hope this helps you!

~Jurgen

6 0

3 years ago

Read 2 more answers

A regular m-gon has 3 times as many sides as a regular k-gon.

Novosadov [1.4K]

9514 1404 393

Answer:

k = 10
m = 30

Step-by-step explanation:

The interior angle of the m-gon is ...

m-interior = 180 -360/m

The exterior angle of the k-gon is ...

k-exterior = 360/k

The required relationships are ...

m = 3k

m-interior = 14/3(k-exterior)

Substituting for m, we can write the latter relation as ...

(180 -360/(3k)) = 14/3(360/k)

Multiplying by 3k/180, we have ...

3k -2 = 28

k = (28 +2)/3 = 10

The values of k and m are 10 and 30, respectively.

_____

<em>Check</em>

The interior angle of the m-gon is 180 -360/30 = 168 degrees.

The exterior angle of the k-gon is 360/10 = 36 degrees.

The angle ratio is 168/36 = 14/3 as required.

3 0

3 years ago

use the accompanying statcrunch session to obtain a simple random sample of 5 high schools in the city of chicago. loading... cl

svetoff [14.1K]

Using sampling concepts, one sample of 5 high schools in the city of Chicago is given by:

Brooks, Hyde Park, Young Womens, Marshall, Noble - UIC.

<h3>What is the missing information?</h3>

The problem is incomplete, but researching it on a search engine, it gives us a list of 15 high schools in Chicago, and asks us to take a sample of 5.

<h3>What is population and sample?</h3>

Population: Collection or set of individuals or objects or events whose properties will be studied.

Sample: The sample is a subset of the population, and a well chosen sample, that is, a representative sample will contain most of the information about the population parameter. A representative sample means that all groups of the population are inserted into the sample.

Hence, from the concepts of sample and populations, the sample of 5 means that we have to select 5 schools from the 15 listed, hence one possible option is given by:

Brooks, Hyde Park, Young Womens, Marshall, Noble - UIC.

More can be learned about sampling concepts at brainly.com/question/25122507

#SPJ1

7 0

2 years ago

If sin theta = (4)/(7), theta in quadrant II, find the exact value of (a) cos theta (b) sin (theta + (pi) / (6) ) (c) cos (the

EleoNora [17]

Answer:

a) $\cos(\theta) = \frac{\sqrt[]{33}}{7}$

b) $\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}$

c) $\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}$

d) $\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

Step-by-step explanation:

We will use the following trigonometric identities

$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$

$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$ $\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$ .

Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that

$x^2+4^2 = 7 ^2$

which implies that $x=-\sqrt[]{49-16} = -\sqrt[]{33}$ . Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then

$\cos(\theta) = \frac{-\sqrt[]{33}}{7}$

b)Recall that $\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}$ , then using the identity from above, we have that

$\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}$

c) Recall that $\sin(\pi)=0, \cos(\pi)=-1$ . Then,

$\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}$

d) Recall that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}$ . Then

$\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

5 0

3 years ago

What is a(n) = -6-4 (n-1)

Ronch [10]

i think this is the answer for n=-2/a+4

3 0

4 years ago