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3 years ago

A circular mirror has a diameter of 12 inches. part a what is the area, in square inches, of the mirror?

Mathematics

1 answer:

vovangra [49]3 years ago

6 0

1st part is c
2nd part is d

I BELIEVE

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Solve the equation: (x^4 - x + y)dx - xdy=0

algol [13]

Keeping in mind that "dx" is the delta x, or differential over the x-axis, whilst "dy" is the delta y or differential over the y-axis,

$\bf (x^4-x+y)dx-xdy=0\implies \stackrel{\textit{distributing dx}}{x^4dx-xdx+ydx}-xdy=0
\\\\\\
ydx=xdy-x^4dx+xdx\implies y=\cfrac{xdy-x^4dx+xdx}{dx}
\\\\\\
y=\stackrel{\textit{distributing the denominator}}{x\cfrac{dy}{dx}-x^4+x}$

5 0

3 years ago

If the hypotenus is 41 inches, and one of the legs is 40 inches what is the length of the other Leg

andre [41]

You find the answer through the Pythagorean theorem

a^2 + b^2 = c^2 a & b are the legs and c is the hypotenuse
a = 40
b = ?
c = 41

40^2 + b^2 = 41^2
1600 + b^2 = 1681
b^2 = 1681 - 1600
b^2 = 81
b = 9

the length of the other leg is 9

i hope you found this helpful!

7 0

3 years ago

Which graph represents f(x)=−2cos(4πx)?

algol [13]

Answer:

the last one (bottom right)

Step-by-step explanation:

4×pi is the same as twice around the whole circle to the beginning. it is the same as 0°, and cos(0) = 1, and then -2×cos(0) = -2

so, we only need to look, where the function reaches -2.

it should be at x = 0, x = 0.5 and x = 1.

3 0

2 years ago

a. Among a shipment of 5,000 tires, 1,000 are slightly blemished. If one purchases 10 of these tires, what is the probability th

hodyreva [135]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

3 or less than 3 tires among the 10 tires should be blemished.

There are some possibilities.

1000 tires are blemished, (5000 - 1000) = 4000 tires are not blemished.

From the 5000 tires, 10 tires can be chosen in $^{5000}C_{10}$ ways.

Possibility 1: 3 tires are blemished.

Total 10 tires can be chosen with 3 blemished tires in $^{1000}C_3 \times {4000}C_7$ ways.

Possibility 2: 2 tires are blemished.

Total 10 tires can be chosen with 2 blemished tires in $^{1000}C_2 \times ^{4000}C_8$ ways.

Possibility 3: 1 tires are blemished.

Total 10 tires can be chosen with 1 blemished tires in $^{1000}C_1 \times^{4000}C_9$ ways.

Possibility 4: No tires are blemished.

Total 10 tires can be chosen with no blemished tires in $^{1000}C_0 \times^{4000}C_{10}$ ways.

Hence, the required probability is $\frac{^{1000}C_1 \times^{4000}C_9 + ^{1000}C_0 \times^{4000}C_{10} + ^{1000}C_2 \times^{4000}C_8 + ^{1000}C_3 \times^{4000}C_7}{^{5000}C_{10} }$ .

6 0

3 years ago

Mathematics can I have a hand.... :´)

iogann1982 [59]

Answer: I honestly don’t know

Step-by-step explanation: I’m so sorry

4 0

3 years ago