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3 years ago

How many times larger is “8 x 10 to the power of 4” then “8 x 10 to the power of 1”

Mathematics

2 answers:

Pachacha [2.7K]3 years ago

3 0

The Answer would be D

wolverine [178]3 years ago

3 0

Answer: D

I hope this helps you. :)

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Yoo i need help !!! its math and i havent learned it yet

BlackZzzverrR [31]

Answer:

${f}^{ - 1} (x) = 9x + 18$

Step-by-step explanation:

$f(x) = \frac{1}{9}x - 2$

$y = \frac{1}{9} x - 2$

$x = \frac{1}{9}y - 2$

$x + 2 = \frac{1}{9}y$

$9x + 18 = y$

${f}^{ - 1} (x) = 9x + 18$

4 0

3 years ago

Read 2 more answers

B= c= Please help asap

gladu [14]

B= 20°
and c = 160°

b is equal to the given angle, b = 20°

c is 180 - 20 = 160°

4 0

2 years ago

Give me examples of equivalent expressions of square root 54 with a radican of 3

stepan [7]

Try this answer 3√54

8 0

2 years ago

A pan is heated to 393°F, then removed from the heat and allowed to cool in a kitchen where the room temperature is a constant 6

deff fn [24]

Answer:

The approximate temperature of the pan after it has been away from the heat for 9 minutes is 275.59°F.

Step-by-step explanation:

The formula for D, the difference in temperature between the pan and the room after t minutes is:

$D = 324\cdot e^{-0.05t}$

Compute the approximate difference in temperature between the pan and the room after 9 minutes as follows:

$D = 324\cdot e^{-0.05t}$

$=324\times e^{-0.05\times 9}\\\\=206.59$

Then the approximate temperature of the pan after it has been away from the heat for 9 minutes is:

D = P - R

206.59 = P - 69

P = 206.59 + 69

P = 275.59°F

Thus, the approximate temperature of the pan after it has been away from the heat for 9 minutes is 275.59°F.

6 0

2 years ago

From a practice assignment:<br>solve the following differential equation given initial conditions

hodyreva [135]

If $y' = e^y \sin(x)$ and $y(-\pi)=0$ , separate variables in the differential equation to get

$e^{-y} \, dy = \sin(x) \, dx$

Integrate both sides:

$\displaystyle \int e^{-y} \, dy = \int \sin(x) \, dx \implies -e^{-y} = -\cos(x) + C$

Use the initial condition to solve for $C$ :

$-e^{-0} = -\cos(-\pi) + C \implies -1 = 1 + C \implies C = -2$

Then the particular solution to the initial value problem is

$-e^{-y} = -\cos(x) - 2 \implies e^{-y} = \cos(x) + 2$

(A)

4 0

1 year ago