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4 years ago

Rewrite each expression using the base only once 1) 7^3 x 7^4

1 answer:

8 0

1) 7^3 x 7^4 = 7^(3+4) = 7^7

2) (-6)^12 x (-6)^5 x (-6)^2 = (-6)^(12+5+2) = (-6)^19

3) 2^2 x 2^7 x 2^ 0 = 2^(2+7+0) = 2^9

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Need help please. I don't think this is dimensionally correct however I want to be sure. Please check if s = (2)(u+v)t^2 is dime

irga5000 [103]

Your equation is not dimensionally consistent.

Let L = length and T = time. Without taking into account precise units, you can write velocity as the ratio between some distance L and some time T.

So in terms of these fundamental quantities, your equation can be written as

L = (L/T + L/T)T^2

L = (L + L)T

L = LT

7 0

4 years ago

PLEASE ANSWER ASAP!!!!

Arlecino [84]

We'll use the notation $\textrm{Arcsin } x$ for the principal value and $\arcsin x$ for all the values:

$\arcsin x = \textrm{Arcsin } x + 2\pi k \textrm{ or } (\pi - \textrm{Arcsin }x) + 2\pi k \quad \textrm{ integer } k$

Part I

$\sin(\theta/2) = \frac 1 2$

$\theta/2 = \arcsin \frac 1 2$

Part II. In our range we can write

$\sin(\theta/2) = \sin(\frac \pi 6)$

$\theta/2 = \frac \pi 6 \textrm{ or } \theta /2 = \pi - \frac \pi 6 = \frac{5\pi}{6}$

Part III.

$\theta = \frac \pi 3 \textrm{ or } \theta = \frac{5\pi}{3}$

Part IV.

$\theta/2 = \frac \pi 6 + 2\pi k \textrm{ or } \theta = \frac{5 \pi}{6} + 2\pi k \quad \textrm{integer } k$

$\theta = \frac \pi 3 + 4 \pi k \textrm{ or } \theta = \frac{5 \pi}{3} + 4\pi k \quad \textrm{integer } k$

5 0

3 years ago

Ssessment

Ostrovityanka [42]

Answer:

Step-by-step explanation:

1/2 bh

1/2 x 7 x 9

= 31.5 cm^2

7 0

3 years ago

An Uber fare is calculated using the formula: Base Fare + (Cost per minute x time of the ride in hours).

pantera1 [17]

Answer:

$40

Step-by-step explanation:

substitute numbers:

20 + (10x3)

20+30= 40.

3 0

2 years ago

Pls pls help! 50pts!!!

Alchen [17]

Answer:

I figured all these out by graphing the function. See the attachment for a picture.

h(x) = (x-1)^2

It's increasing over (1, ∞) because the values are going up in that span.

It's decreasing over (-∞, 1) because the values are going down in that span.

Its range is all values of y $\geq$ 0 (we can see that only y-values above 0 are in the function).

g(x) = -(x+1)(x-1)

It's increasing over (-∞, 0) because the values are going up in that span.

It's decreasing over (0, ∞) because the values are going down in that span.

Its range is all values of y $\leq$ -1 (we can see that only y-values below 1 are in the function).

5 0

3 years ago

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