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2 years ago

Please help (images below)

Mathematics

1 answer:

Kay [80]2 years ago

5 0

Answer:

2) D

because:

$\frac{9}{1} \times 4 \\ 9 \times 4 \\ 36 \: is \: not \: equal \: to \: 4$

4) B

because:

$\frac{5}{5} \times 8 \\ 1 \times 8 \\ 8 \: is \: equal \: to \: 8$

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A baker uses 4 berries for every muffin which equation can be used to find the number of berries b needed to make m muffins

KATRIN_1 [288]

Answer:

b = 1/4m

Step-by-step explanation:

A baker uses 4 berries for every muffin which equation can be used to find the number of berries b needed to make m muffins

Given :

Berries = b

Muffins = m

4 berries for every muffin

4b = m - - - (1)

Th number of berries to make m muffins:

Make b the subject :

Divide both sides by 4

b = m/4

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3 years ago

4 yd 4 yd 20 yd 24 yd 20 yd 24 yd What is the perimeter?

Mice21 [21]

Answer:

Step-by-step explanation:

24+20+24+20+4+4=96

7 0

4 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

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

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