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3 years ago

10

What two mixed numbers between 3 and 4 have a product of 9 and 12

1 answer:

dlinn [17]3 years ago

8 0

It's a trick question. There are an infinite number of mixed numbers between 3 and 4 that can multiply to equal 12 (for example, 3 and 3/7 times 3 and 1/2), but there are no mixed numbers between 3 and 4 that can multiply to equal 9. 3 times 3 is not between them but is 3, but that quantity is excluded because 3<x<4. Anything even a small bit above the number 3 would have to be multiplied by 2 and some fraction, which would not be between 3 and 4.

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6-2x=5x-9x+10 solve for x

andrew-mc [135]

X=2

So you solve for x by simplifying bath sides of the equation, then all you do is isolate the variables

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2 years ago

Given g(x) = :-2 +4, solve for a when g(x) = 0.

evablogger [386]

Answer:

-7/4

Step-by-step explanation:

You are looking for the composite g(f(2)). The simplest way to solve this is to evaluate f(2) and enter the solution in to your g function.

g(f(2))=g(-(2)^2-2(2)+4)=g(-4-4+4)=g(-4)

g(-4)=4/(-4(-4)-2)=4/(16-2)=4/14=2/7

Therfor, g(f(2))=2/7 **I'm assuming the -4x-2 is all in the denominator of the g(x) function. If -2 is not in the denominator you would have

g(f(2))=4/(-4(-4)) -2=4/16 -2=1/4 -2=1/4-8/4= -7/4

8 0

2 years ago

If town A had 35 people, 12 left, 13 joined, 39 joined, 25 got killed, how many people were there left?

Makovka662 [10]

Answer:

50

Step-by-step explanation:

35 - 12 + 13 + 39 - 25 = 50

6 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csec%5Cleft%28x%5Cright%29%7D%7B%5Ccos%5Cleft%28x%5Cright%29%7D-%5Cfrac%7B%5Csin%5

DanielleElmas [232]

Answer:

1

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$ .

Thus:

$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

$=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }$

Let's do the first part first: (Recall how to divide fractions)

$\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}$

For the second term:

$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

$\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}$

Note the numerator; it can be derived from the Pythagorean Identity:

$sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)$

Thus, we can substitute the numerator:

$\frac{1-sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)}=1$

Everything simplifies to 1.

7 0

2 years ago

Are y = 1/4 -2 and y = 1/4 + 2 parallel?

zhenek [66]

Answer:

yes because the slope is equal. so these two lines are parallel.

3 0

2 years ago