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3 years ago

Create a story problem for 3/4 times 3

Mathematics

1 answer:

luda_lava [24]3 years ago

8 0

Answer:

Kendrick is making a buffet for his family. He needs 3x3/4 of flour to make it.

He doesn't know what store to go to either.

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A tree is leaning at a 5° angle from vertical, making an 85° angle with the ground. When the sun is at a 65° angle of elevation,

Vinvika [58]

D = 25 ft is the length of a shadow. L - the length of a tree.
Two angles are 85° and 65° and the third is 180° - ( 65° + 85° ) =
= 180° - 150° = 30°.
We will use the Sine Law:
25 / sin 30° = L / sin 65°
25 / 0.5 = L / 0.9063
25 * 0.9063 = 0.5 L
22.6577 = 0.5 L
L = 22.6577 : 0.5
L = 45.3 ft.
Answer: the approximate length of the tree is 45.3 ft.

3 0

3 years ago

Len [333]

Answer:

isp2odbskoxnxndkekjrhsbnzmcnfjhdhdhhdhdhshjsh

7 0

3 years ago

What is the interquartile range of the following data set?

Brilliant_brown [7]

Pretty sure the answer is A

6 0

3 years ago

ASAP!!!! Please help me. This is due in 10 minutes. Find the volume

Karo-lina-s [1.5K]

Answer:

The answer is 960.

8 x 12 x 20 is 1920. 1920 divided by 2 is 960.

6 0

3 years ago

Read 2 more answers

1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

german

Answer:

Step-by-step explanation:

To write the form of the partial fraction decomposition of the rational expression:

We have:

$\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}$

Using partial fraction decomposition to find the definite integral of:

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx$

By using the long division method; we have:

$x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }$

$- 2x^3 -16x^2-40x$

$x+ 20$

So;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}$

By using partial fraction decomposition:

$\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}$

$= \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}$

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now; we have to relate like terms on both sides; we have:

A + B = 1 ; 2A - 10 B = 20

By solvong the expressions above; we have:

$A = \dfrac{5}{2}$ $B = \dfrac{3}{2}$

Now;

$\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}$

Thus;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}$

Now; the integral is:

$\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx$

$\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}$

3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.

Please check to the attached image below for the solution to question number 3.

4 0

3 years ago