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emmainna [20.7K]

3 years ago

8

Is point F interior to both circles

1 answer:

faltersainse [42]3 years ago

8 0

answer: point F is inferior to both circles

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I roll a standard six-sided die 6 times in a row. What is the probability that the numbers

ioda

I think 50% or 1/2 ( it’s the same just one is in fraction and other in percentage)

7 0

3 years ago

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Jada typically earns$1545 each month, of which $47.20 is spent on electricity. Whick percent of jadas earnings are spent on elec

REY [17]

The equation is asking how much out of Jada's total budget (1545) is spent on electricity a month (47.20).
write this as a fraction 47.20/1545
type into calculator and you get .03055
multiply that by 100 to get it as a percent. you get 3.055%
now round that to the nearest tenth and you get 3.1%
answer 3.1%

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3 years ago

Divide for 2class .45÷6789

FrozenT [24]

<span>0.00006628369 is your answer for 0.45/6789</span>

8 0

3 years ago

max2010maxim [7]

Hello!

So, this is quite the complex question, and here are the following steps:

What is the quotient of $\frac{6-3\sqrt[3]{6}}{\sqrt[3]{9}}$ ?

$\frac{(6 - 3\sqrt[3]{6})\sqrt[3]{9^{2}}}{9}$ (rationalize the denominator)

$\frac{3(2 -\sqrt[3]{6})\sqrt[3]{9^{2}}}{9}$ (factor 3 from the expression)

$\frac{(2-\sqrt[3]{6})\sqrt[3]{9^{2}}}{3}$ (reduce the fraction with 3)

$\frac{2\sqrt[3]{9^{2}}-\sqrt[3]{9^{2}}}{3}$ (distributive property)

$\frac{2\sqrt[3]{81}-\sqrt[3]{486}}{3}$ (simplify 3 · 9²)

$\frac{6\sqrt[3]{3}-3\sqrt[3]{18}}{3}$ (simplify the radical)

$\frac{3(2\sqrt[3]{3}-3\sqrt[3]{18})}{3}$ (factor 3 from the expression)

$2\sqrt[3]{3}-\sqrt[3]{18}$ (reduce the fraction)

The answer, is simply, choice A, $2\sqrt[3]{3} -\sqrt[3]{18}$ ≈ 0.263758.

5 0

3 years ago

Two semicircles are attached to the sides of a rectangle as shown.

melomori [17]

Answer:

$157\ in^{2}$

Step-by-step explanation:

we know that

The area of the figure is equal to the area of rectangle plus the area of two semicircles

<u>The area of rectangle is equal to</u>

$A=14*5=70\ in^{2}$

<u>The area of the small semicircle is equal to</u>

$A=\pi r^{2} /2$

$r=5/2=2.5\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(2.5^{2})/2=9.8125 in^{2}$

<u>The area of the larger semicircle is equal to</u>

$A=\pi r^{2} /2$

$r=14/2=7\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(7^{2})/2=76.93\ in^{2}$

The area of the figure is equal to

$70+9.8125+76.93=156.7425=157\ in^{2}$

8 0

3 years ago

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