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velikii [3]
2 years ago
12

What is the solution to the division problem below? (You can use long division or synthetic division) X^3 + x^2 - 11x + 4/x + 4

Mathematics
2 answers:
Tom [10]2 years ago
4 0

Answer:

B. x2-3x+1

Step-by-step explanation:

mel-nik [20]2 years ago
3 0

Answer:

X2-3x+1

Step-by-step explanation:

B RAINLY

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In a class there are 40 students 30 are boys and what is the percentage of girls ....​
Artyom0805 [142]

Answer:

25%

Step-by-step explanation:

Total= 40 students

boys +girls= 40

30 +girls= 40

number of girls

= 40 -30

= 10

Percentage of girls

=  \frac{10}{40}  \times 100\% \\ =  25\%

Alternatively,

percentage of girls

= 100% -(percentage of boys)

= 100\% - ( \frac{30}{40}  \times 100\%) \\  = 100\% - 75\% \\  = 25\%

6 0
3 years ago
0.6 litres in the ratio 7:5
Sedaia [141]
Can you Elaborate your question?
5 0
2 years ago
Triangle PQR was dilated according to the rule DO,2(x,y)(2x,2y) to create similar triangle P'Q'Q. On a coordinate plane, (0, 0)
Readme [11.4K]

Answer:

1 and 4

Step-by-step explanation:

Ed2020

8 0
3 years ago
Read 2 more answers
Franklin uses the distributive property to write an expression equivalent to 18 + 36. Which of the following best represents Fra
allsm [11]
18+36=54
i tried to do all of these and none of them add up to 54
6 0
3 years ago
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A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line
Nataliya [291]

Answer:

Volume is 2000\pi\ m^{3}

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d              (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d                  (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

c = \frac{1}{5}

Therefore,

h(y) = \frac{1}{5}y + 5

Now, the Volume of the pool is given by:

V = \int h(y)dA

where

A = r\theta

A = rdr\ d\theta

Thus

V = \int (\frac{1}{5}y + 5)dA

V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta

V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta

V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta

V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}

V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}

7 0
3 years ago
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