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GenaCL600 [577]

3 years ago

9

Simplify x^2(y^3)^4/xy^5=x^a•y^b

1 answer:

givi [52]3 years ago

7 0

Answer: a = 1 b = 7

Step-by-step explanation:

x^2(y^3)^4/x.y^5 = x^a.y^b

x^2(y^12) = x^a.y^b

(x^2).(y^12)/x^y^5 = x.y^7

x.y^7= x^a . y^b -> <u>x^1.y^7</u>

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Please answer hurry!!!!

lutik1710 [3]

The best way to solve this problem is by plugging in each answer choice for both equations and seeing which one works. You can already see that choice A doesn't work because -1 is not greater than or equal to 4. That means you just need to test choices B and C!

Choice B (4, -2)
$x + y \geq 6\\
4 + -2 \geq 6\\
2 \geq 6 
$
2 is not greater than or equal to 6, so choice B doesn't work.

Choice C (4, 2)
$x + y \geq 6\\ 
4 + 2 \geq 6\\ 
6 \geq 6 \\\\
x \geq 4\\
4 \geq 4$
Since 6 is equal to 6 and 4 is equal to 4, choice C works!

-------

Answer: C) (4, 2)

6 0

3 years ago

Suppose c was a total membership cost for m the two equations that model c are c=30m+50 an c=20m+100

Maksim231197 [3]

Answer:

5 months

Step-by-step explanation:

Given

$c = 30m + 50$

$c = 20m + 100$

Required [Missing from the question]

At what month will the cost be equal

To do this, we equate both expressions

$c = c$

This gives:

$30m + 50 = 20m + 100$

Collect like terms

$30m - 20m = 100 - 50$

$10m = 50\\$

Solve for m

$m = \frac{50}{10}$

$m = 5$

8 0

3 years ago

Find two z values, one positive and one negative, that are equidistant from the mean so that the areas in the two tails add to t

levacccp [35]

I think it would be the answer is b.

3 0

4 years ago

Factor completely: 3x^3+33x^2+54x=0

kap26 [50]

Answer:

3(x + 9)(x + 2)x = 0

Step-by-step explanation:

xv1 = −9

xv2 = −2

xv3 = 0

7 0

3 years ago

What is f–1(8)??<br><br> Please select the best answer

notka56 [123]

Answer:

$f^{-1}(8)=\frac{3}{2}$

Step-by-step explanation:

Given f(x)= $2x+5$

We have to find $f^{-1}(8)$

In order to find $f^{-1}(x)$ we have to make x as the subject of the formula

Let us assume f(x)=y

⇒ $2x+5=y$

Subtracting both sides by 5

$2x+5-5=y-5$

$2x=y-5$

Dividing both sides by 2

$\frac{2x}{2}=\frac{y-5}{2}$

⇒ $x=\frac{y-5}{2}$

Now substituting y with x

we have $f^{-1}(x)=\frac{x-5}{2}$

Now $f^{-1}(8)=\frac{8-5}{2}$

= $\frac{3}{2}$

So option (ii) is correct

5 0

3 years ago

Read 2 more answers