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4 years ago

Need help on ratio table

Mathematics

1 answer:

guajiro [1.7K]4 years ago

5 0

80.8, 40.4, 20.2, 10.1 (the numbers are going in half.)

1024, 512, 256, 128. (also going in half.)

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Evaluate the expression. simplify your answer. 2/3 x 1/4 / 4/5

Arturiano [62]

Answer:

The answer is 5/24.

Step-by-step explanation:

2/3 multiplied by 1/4 is 2/12 which simplifies to 1/6. CCF (copy, change, flip) is applied to 4/5 and modifies the equation to become 1/6*5/4 for a final answer of 5/24.

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3 years ago

Read 2 more answers

HELPPP PLEASE .☆........

tiny-mole [99]

Answer:

I think its 1/3/ and 4. But I dont know tho. just make sure with other.

Step-by-step explanation:

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3 years ago

Find and x and y <br> PLEASE HELP

Bogdan [553]

Answer:

Step-by-step explanation:

x is 79 and y is 47

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3 years ago

What Makes This Number Unique? 8,549,176,320

Serggg [28]

Answer:

When spelt out the digits are in alphabetical order

eight

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nine

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Step-by-step explanation:

7 0

3 years ago

Let R be the region bounded by

loris [4]

a. The area of $R$ is given by the integral

$\displaystyle \int_1^2 (x + 6) - 7\sin\left(\dfrac{\pi x}2\right) \, dx + \int_2^{22/7} (x+6) - 7(x-2)^2 \, dx \approx 9.36$

b. Use the shell method. Revolving $R$ about the $x$ -axis generates shells with height $h=x+6-7\sin\left(\frac{\pi x}2\right)$ when $1\le x\le 2$ , and $h=x+6-7(x-2)^2$ when $2\le x\le\frac{22}7$ . With radius $r=x$ , each shell of thickness $\Delta x$ contributes a volume of $2\pi r h \Delta x$ , so that as the number of shells gets larger and their thickness gets smaller, the total sum of their volumes converges to the definite integral

$\displaystyle 2\pi \int_1^2 x \left((x + 6) - 7\sin\left(\dfrac{\pi x}2\right)\right) \, dx + 2\pi \int_2^{22/7} x\left((x+6) - 7(x-2)^2\right) \, dx \approx 129.56$

c. Use the washer method. Revolving $R$ about the $y$ -axis generates washers with outer radius $r_{\rm out} = x+6$ , and inner radius $r_{\rm in}=7\sin\left(\frac{\pi x}2\right)$ if $1\le x\le2$ or $r_{\rm in} = 7(x-2)^2$ if $2\le x\le\frac{22}7$ . With thickness $\Delta x$ , each washer has volume $\pi (r_{\rm out}^2 - r_{\rm in}^2) \Delta x$ . As more and thinner washers get involved, the total volume converges to

$\displaystyle \pi \int_1^2 (x+6)^2 - \left(7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \pi \int_2^{22/7} (x+6)^2 - \left(7(x-2)^2\right)^2 \, dx \approx 304.16$ <em />

d. The side length of each square cross section is $s=x+6 - 7\sin\left(\frac{\pi x}2\right)$ when $1\le x\le2$ , and $s=x+6-7(x-2)^2$ when $2\le x\le\frac{22}7$ . With thickness $\Delta x$ , each cross section contributes a volume of $s^2 \Delta x$ . More and thinner sections lead to a total volume of

$\displaystyle \int_1^2 \left(x+6-7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \int_2^{22/7} \left(x+6-7(x-2)^2\right) ^2\, dx \approx 56.70$

7 0

2 years ago