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2 years ago

8

Jean noticed that the tar pit bubbled more on warm days than on cool days or at night. Is the air temperature the dependent vari

able or the independent variable?
A. Independent
B. Dependent

1 answer:

andreev551 [17]2 years ago

6 0

It's indepentent. In fact, the tar pit's temperature DEPENDS (dependent variable) by the air's temperature (independent variable).

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It takes 27 pounds of seed to completely plant a 3-acre field. How many pounds of seed are needed per acre?

Ray Of Light [21]

Answer:

9 pounds of seed per acre.

Step-by-step explanation:

All you have to do is division. It takes 27 pounds of seed to make 3 acres, so 27 divided by 3. That gives you 9.

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2 years ago

Debra start a babysitting business. She purchased toys, snacks, and other supplies. She charged an hourly rate for each child. T

Ghella [55]

Answer:

A. 6y

Step-by-step explanation:

The profit formula is: total earnings- total expenses. In this case the 145 has a minus sign, which indicates that these are Debra's total expenses. Then, 6y represents total revenue. You can also identify that this is the answer because revenues depend on babysitting hours, which in this case, is represented by the variable "y".

8 0

2 years ago

(5,9), (5-3)<br> what is the slope?

monitta

Answer:

Hello!!! Princess Sakura here ^^

Step-by-step explanation:

When you want to find out what the slope is, you have to remember what the formula is first....

$m=\frac{y}{x} =\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

so in this case -3 would be the $y_{2}$ because it's the y in the ordered pair and it's the second one. Then one of the 5's would be $x_{2}$ and 9 would be $y_{1}$ and then the other 5 would be $x_{1}$ .

now if we plug it in

$\frac{-3-9}{5-5} =\frac{-12}{0}$

now we can tell that we can't divide by zero so the slope would be unidentified and that's it.

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2 years ago

Integrate the following

enot [183]

I suppose you mean to have the entire numerator under the square root?

$\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx$

We can use a trigonometric substitution to start:

$x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt$

Then for $x=2$ , $t=\sec^{-1}1=0$ ; for $x=4$ , $t=\sec^{-1}2=\frac\pi3$ . So the integral is equivalent to

$\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt$

We can write

$\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t$

so the integral becomes

$\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}$

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2 years ago

I need help I’ll give brainlist answer.

vodomira [7]

Answer:

All beside D

Step-by-step explanation:

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2 years ago