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Assoli18 [71]
3 years ago
5

HELP ME PLEASE I NEED ANSWERS REALLY FAST I'LL GIVE BRAINLIEST TO THE CORRECT ANSWER SO PLEASE HELP ME!!!!

Mathematics
1 answer:
GuDViN [60]3 years ago
3 0

Answer:

B

Step-by-step explanation:

These lines dont cross CD and are not parrellel which is what skew lines are.

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What is the solution of the proportion? 6/a = 18/27
Marta_Voda [28]

Answer:

a=9

Step-by-step explanation:

To solve this proportion, we have to get the variable, a, by itself.

First, cross multiply.

6/a=18/27

Multiply the denominator of the first fraction by the numerator of the second, and the numerator of the second by the denominator of the first.

a*18=6*27

18a=162

Now, 18 and a are being multiplied. In order to get a by itself, perform the opposite of what is being done. They are being multiplied, so the opposite would be division. Divide both sides by 18.

18a/18=162/18

a=162/18

a=9

So, the proportion, with 9 substituted in for a, will be:

6/9=18/27

6 0
3 years ago
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Given y = 2(x-3) and a domain of (-3,-2 and 5). Find the range.
alexandr402 [8]
The answers are (-3,-12), (-2,-10), (5,4)
4 0
3 years ago
Meghan and Sabrina compared the amount of interest they each earned on their savings accounts. Each had deposited $1000, but Meg
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Sabrina because she earned more money then  meghen
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Ina survey at a supermarket, 78 out of 120 people surveyed said that they shopped more than ounce a week. What percent of the pe
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78 /120 =0.65

0.65 x 100 = 65%

the answer is 65%

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3 years ago
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a) Estimate the volume of the solid that lies below the surface z = 7x + 5y2 and above the rectangle R = [0, 2]⨯[0, 4]. Use a Ri
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In the x direction we consider the m=2 subintervals [0, 1] and [1, 2] (each with length 1), while in the y direction we consider the n=2 subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of R are (1, 0), (2, 0), (1, 2), (2, 2).

Let f(x,y)=7x+5y^2. The volume of the solid is approximately

\displaystyle\iint_Rf(x,y)\,\mathrm dx\,\mathrm dy\approx f(1,0)\cdot1\cdot2+f(2,0)\cdot1\cdot2+f(1,2)\cdot1\cdot2+f(2,2)\cdot1\cdot2=\boxed{164}

###

More generally, the lower-right-corner Riemann sum over m=\mu and n=\nu subintervals would be

\displaystyle\sum_{m=1}^\mu\sum_{n=1}^\nu\left(7\frac{2m}\mu+5\left(\frac{4n-4}\nu\right)^2\right)\frac{2-0}\mu\frac{4-0}\nu=\frac83\left(101+\frac{21}\mu+\frac{40}{\nu^2}-\frac{120}\nu\right)

Then taking the limits as \mu\to\infty and \nu\to\infty leaves us with an exact volume of \dfrac{808}3.

7 0
3 years ago
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