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3 years ago

Which expression is equivalent to 2^-3/2^-5

Mathematics

1 answer:

Aleks04 [339]3 years ago

3 0

Answer:

2^2

Step-by-step explanation:

2^-3/2^-5

We know that a^b / a^c = a^(b-c)

2^-3/2^-5 = 2^( -3 - -5) = 2^(-3+5) = 2^2

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Which equation represents a line that passes through (–9, –3) and has a slope of –6?

tatiyna

Answer:

y+3 = -6(x+9)

Step-by-step explanation:

m(x-x1)= (y-y1)

-6[x-(-9)] = [y-(-3)]

-6(x+9) = y+3

.....

6 0

3 years ago

The formula for the area of a triangle is, 1 А A= bh Calculate the height of a triangle when A= 24 m2 and b = 8 m.

Sphinxa [80]

I hope this helps you

A=1/2.b.h

given Area is 24 and base is 8

24=1/2.8.h

24=4.h (divide 4 both of sides)

h=6 (height)

3 0

3 years ago

I need help on this and the person who answer this correctly gets a BRANLIST(answer if you know)

svp [43]

The answer that you’re looking for is B

3 0

3 years ago

Read 2 more answers

123

ser-zykov [4K]

First, I'm going to separate factor the expression inside of the square root.

sqrt[ (2/18) * (x^5) ]

sqrt[ (1/9) * (x^5) ]

We can take the square root of 1/9 easily, because 1 and 9 are both perfect squares. The square root of 1/9 is 1/3.

Looking at the x^5, we can separate it into x^2 * x^2 * x^1. The square root of x^2 is x. So, we now also have an x^2 on the outside because there are two x^2s in our expanded form.

ANSWER: (x^2 * sqrt(x)) / 3

(Option 1)

Hope this helps!

4 0

3 years ago

Ocean sunfish are well known for rapidly gaining a lot of weight on a diet based on jellyfish. The relationship between the elap

Maksim231197 [3]

Answer:

Every day, the mass of the sunfish is multiplied by a factor of [ln(1.34)/6].

Step-by-step explanation:

You have the following function:

$M(t)=(1.34)^{\frac{t}{6}+4}$

To know what is the factor that multiplies the mass of the sunfish each day, you derivative the function M(t):

$\frac{dM(t)}{dt}=(1.34)^{\frac{t}{6}+4}(\frac{1}{6})ln(1.34)\\\\\frac{dM(t)}{dt}=\frac{ln(1.34)}{6}[(1.43)^{\frac{t}{6}+4}]\\\\\frac{dM(t)}{dt}=\frac{ln(1.34)}{6}M(t)$ (1)

where you have used the following general derivative:

$g(t)=a^{f(t)}\\g'(t)=a^{f(t)}f'(t)ln(a)$

The derivative give you the increase in the mass per day (because t is days). By the expression (1) you can conclude that each day the mass increase a factor of [ln(1.34)/6].

Every day, the mass of the sunfish is multiplied by a factor of [ln(1.34)/6].

8 0

3 years ago