All categories

4 years ago

Which expressions are equivalent to 2^5•2^4

Mathematics

2 answers:

muminat4 years ago

7 0

2^9 is the same as your question

labwork [276]4 years ago

3 0

2^5 * 2^4 = 2^(5 + 4) = 2^9

To multiply powers with the same base, add the exponents.

You might be interested in

A rectangle is 12 feet long and 5 feet wide. If the length of the rectangle is increased by 25% and the width is decreased by 20

Evgen [1.6K]

Answer:

no change in area

Step-by-step explanation:

The original area is

A = 12*5 = 60 ft^2

The new length and width

l = 12 + .25 (12) = 12+3 =15

w = 5 - .2 (5) =5-1 = 4

The new area is

A = l*w =15*4 = 60 ft^2

The area is the same

7 0

3 years ago

$4.95 chocolate bars are now on sale for 35% off. what will the chocolate bars cost now? (show work)

Tema [17]

= 4.95 X .65 = 3.2175 ( 3.22).
4.95 = 100% of the price
100 less 35 = 65
4.95 X .65% = 3.2175 or 3.22

6 0

4 years ago

N - 9.02 = 3.87<br> what is (n)?

Nadya [2.5K]

Answer:

n = 12.89

Step-by-step explanation:

Add 9.02 + 3.87

7 0

3 years ago

Image of parabola with a maximum value at (1, 9), y-intercept at (0, 8) and x-intercepts at (2, 0) and (4, 0).

natka813 [3]

Answer:

(a) 4 seconds

Step-by-step explanation:

The given point (4, 0) on the parabola tells you the answer to the question: the coin was in the air 4 seconds before its height is 0. It is no longer in the air at that point.

<em>Additional comment</em>

In order to match the rest of the problem statement, the first given x-intercept must be (-2, 0).

4 0

2 years ago

Simplify: cos2x-cos4 all over sin2x + sin 4x

GrogVix [38]

Answer:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

Step-by-step explanation:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}$

Apply formula:

$\cos\left(A\right)-\cos\left(B\right)=-2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$ and

$\sin\left(A\right)+\sin\left(B\right)=2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$

We get:

$=\frac{-2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\sin\left(\frac{2x-4x}{2}\right)}{2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\cos\left(\frac{2x-4x}{2}\right)}$