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umka2103 [35]
3 years ago
15

What is the volume of the right rectangular prism?

Mathematics
2 answers:
Kamila [148]3 years ago
7 0
What is 7x5x4, it is 140 so the volume is 140in^3 its a pretty simple problem
boyakko [2]3 years ago
3 0
It’s 140in pretty simple sorry if I did not help you hope it did
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Kiana is staying on the 5th floor of a hotel. She rode an elevator down 7 floors from her room to the parking garage to get her
Serggg [28]

Answer:

the 2nd and 4th expressions are the appropriate ones.  

Step-by-step explanation:

both result in '0' which means that kiana is on the ground floor of the building at the end.

4 0
3 years ago
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In the point (3,-2), what is<br> the x-value? Which<br> quadrant is this point<br> located?
natali 33 [55]

Answer:

Step-by-step explanation:

The x-value is 3. Coordinate pairs are labeled as (x, y). In this case, the 3 is in the x position.

The point is located in the 4th quadrant. Quadrant labels move counter-clockwise, from 1 to 4 as shown below. In Quadrant 4, the x-value will always be positive, and the y-value will always be negative. This point has a positive x-value and negative y-value, so it is in this quadrant.

5 0
1 year ago
a recipe calls for two and two fourth cups of raisins,but Julie only has a one fourth cup measuring cup.how many one fourth cups
Olegator [25]
I believe it would take 10 cups. The reason being 4 cups of 1/4 equals 1 so double it to 8 and add 2 more cups to make 2/4. So add 8 plus 2 and get 10.
4 0
3 years ago
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Find the 97th term of the arithmetic sequence 17, 26, 35, ...
vagabundo [1.1K]

Answer:

881

Step-by-step explanation:

Arithmetic sequences can be written as:

a_n = a_1 + (n-1)d\\a_1 \: is \: 17\\a_n = 17 + (n-1)d\\

The second term is 26,

26 = 17 + (2-1)d\\9 = 1d\\d = 9\\a_n = 17 + (n-1)*9\\a_n = 17 + 9n - 9\\a_n = 8 + 9n

So, the 97^t^h term is

a_9_7 = 8 + 9(97)\\a_9_7 = 8 + 873\\a_9_7 = 881

5 0
3 years ago
Question part points submissions used use newton's method with the specified initial approximation x1 to find x3, the third appr
serious [3.7K]

Set f(x)=2x^3-3x^2+2. Find the tangent line \ell_1(x) to f(x) at the point when x=x_1:

f'(x)=6x^2-6x\implies f'(x_1)=12 (slope of \ell_1)

\implies\ell_1(x)=12(x-x_1)+f(x_1)=12(x+1)-3=12x+9

Set x_2=-\dfrac9{12}, the root of \ell_1(x). The tangent line \ell_2(x) to f(x) at x=x_2 has slope and thus equation

f'(x_2)=\dfrac{63}8\implies\ell_2(x)=\dfrac{63}8\left(x+\dfrac9{12}\right)-\dfrac{17}{32}=7x+\dfrac{151}{32}

which has its root at x_3=-\dfrac{151}{224}\approx-0.6741.

(The actual value of this root is about -0.6777)

5 0
3 years ago
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