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3 years ago

Fórmula para sacar el área de un circulo

Mathematics

1 answer:

saul85 [17]3 years ago

8 0

<span>Fórmula para sacar el área de un circulo:

$A = \pi r^2$
</span>

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A 4-- pounds bag of chocolate bars cost $16.24. What is the unit price

ra1l [238]

Answer:

4.06 i think

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2 years ago

Write and solve a real world problem involving the multiplication of a fraction and a whole number whose product is between 10 a

Fed [463]

A runner ran 1/2 of a mile for 24 days. How many miles did the runner run in all? (1/2×24=12) Therefore, the runner ran 12 miles in all in the course of 24 days.

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3 years ago

NEED ANSWER ASAP!!! Express your answer in scientific notation. 9.3 X 107 - 3.14 X 106

kirza4 [7]

Answer:

$89.86\times 10^6$

Step-by-step explanation:

$9.3\times 10^7 - 3.14\times 10^6\\$

In order to simplify the above expressions we make sure that we get same powers of 10 for both the terms.

So, we multiply and divide the first term with 10

Dividing first term by 10.

$\frac{9.3\times 10^7}{10}\\9.3\times 10^{7-1}\\9.3\times 10^6\\$

Now multiplying it by 10.

$10\times9.3\times 10^6\\93\times 10^6$

Evaluating the new expressions.

$93\times 10^6 - 3.14\times 10^6\\$

Taking $10^6$ common factor out.

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3 years ago

Find the equation of a line that passes through the point (4,2) that is perpendicular to the line y=4/3 x

Mrac [35]

Answer:

$y=\frac{3}{4}x-1$

Step-by-step explanation:

Since we're looking at perpendicular lines, the slope of the new line is the reciprocal of the original line.

So the slope of our new line is $\frac{3}{4}$

So far, we are looking at $y=\frac{3}{4} x+b$

Now we need to find the b value by plugging in the given point.

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So the final equation is $y=\frac{3}{4}x-1$

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3 years ago

A ball is launched straight up in the air from a height of 6 feet. Its velocity (feet/second) t seconds after launch is given by

german

Step-by-step explanation:

A ball is launched straight up in the air from a height of 6 feet. The velocity as a function of time t is given by :

f(t) = -32 t+285

Height of the ball is :

$h(t)=\int\limits{f(t){\cdot} dt}\\\\h(t)=\int\limits{(-32t+285){\cdot} dt}\\\\h(t)=-16t^2+285t+C$

C is constant. Here the ball is launched from a height of 6 feet. So,

$h(t)=-16t^2+285t+6$

At t = 2 s, $h(t)=-16(2)^2+285(2)+6=512\ m$

At t = 9 s, $h(t)=-16(9)^2+285(9)+6=1275\ m$

Between 2 s and 9 s, the ball's height changed is : 1275 - 512 = 763 m.

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3 years ago