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3 years ago

Simple interest. $563,2.5%, 3 years

Mathematics

1 answer:

ella [17]3 years ago

6 0

Yea I tink itss 7000 dollas ya

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A triangle with an angle measuring 140° must be which of these? isosceles equilateral right obtuse acute

sp2606 [1]

Answer:

Step-by-step explanation:

OBTUSE......

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3 years ago

Determine the quotient of 1 and 2 over 3 divided by 4 over 5 . (2 points)

saw5 [17]

Answer:

2 and 1 over 12

Step-by-step explanation:

$1 \frac{2}{3} \div \frac{4}{5} \\ \\ = \frac{1 \times 3 + 2}{3} \div \frac{4}{5} \\ \\ = \frac{5}{3} \times \frac{5}{4} \\ \\ = \frac{25}{12} \\ \\ = 2 \frac{1}{12} \\$

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3 years ago

Karen made $288 for 18 hours of work,

Zinaida [17]

Answer:

She would make 500,000 or else she will speak to the manager.

Step-by-step explanation: Karen will speak to her manager and torture her manager until the manager gives her the money.

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3 years ago

Read 2 more answers

I need help with all of Wednesday and Thursday I'm giving 20 points if your sure your answers are correct thanks for helping

lukranit [14]

3
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im sorry i dint know some of them but i hope this helps

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3 years ago

Read 2 more answers

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the re

Thepotemich [5.8K]

Answer:

Step-by-step explanation:

From the first image attached below, we will see the sketch of the curve x = y² & x = 2y

In the picture connected underneath, the concealed locale(shaded region) is bounded by the given curves. Now, we discover the marks of the crossing point of the curves. These curves will cross, when:

$y^2 =2 y \\ \\ or y^2 -2y = 0 \\ \\ y(y-2) = 0 \\ \\ y = 0 \ \ or \ \ y = 2$

Thus, the shaded region fall within the interval 0 ≤ y ≤ 2

Now, from the subsequent picture appended we sketch the solid acquired by turning the concealed region about the y-axis.

For the cross-sectional area of the washer:

$A (y) = \pi (outer \ radius)^2 - \pi ( inner \ radius )^2 \\ \\ A(y) = \pi (2y^)2- \pi (y^)^2 \\ \\ A(y) = 4 \pi y^2 - \pi y^4 \\ \\ A(y) = \pi( 4 y^2 -y^4)$

Finally, the volume of (solid) is:

$V = \int^2_0 A(y) \ dy \\ \\ V = \int^2_0 \pi (4y^2 -y^4) \ dy \\ \\ V = \pi \int^2_0 (4y^2 -y^4) \ dy \\ \\ V = \pi \Big[\dfrac{4}{3}y^3 - \dfrac{y^5}{5} \big ] ^2_0 \\ \\ V = \pi \Big [ \dfrac{4}{3}(2)^3-\dfrac{2^3}{5} \Big ] \\ \\ V = \dfrac{64}{15}\pi \\ \\ V = (4.27 ) \pi$

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3 years ago