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3 years ago

The radius of a circle is 5m Find it’s area to the NEAREST TENTH

Mathematics

1 answer:

Yanka [14]3 years ago

5 0

Answer:

$78.5 \ m^2$

Step-by-step explanation:

$A=\pi r^2$

Or area equals pi times radius squared

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Giving brainiest+points need explanation on this one my answer is 25 is it right?

natka813 [3]

Known facts:
A line shaft rotating at 250 rpm, a grinder shaft that turns at 1200 rpm diameter of grinder pulley is 5 inches.

1. Proportion equation
350/1200 = 5/x

2. Cross multiply
1200*5=250*x
x=(1200*5)/250 = (24*50*5)/(5*50) = 24
The big wheel diameter is 24 inches

8 0

3 years ago

Help asap plsplspls!! :)

OlgaM077 [116]

Answer:

1.7

Step-by-step explanation:

1 + 7/10

= 1 + 0.7

= 1.7

6 0

3 years ago

Read 2 more answers

4 yd 2 ft = _____ ft. Explain how you knew how to solve the problem. * 2

morpeh [17]

Hi there!

4yds and 2ft are equaled to 14ft.

Each yard consists of 3 feet. If there are 4 then you multiply 4 times 3, resulting in 12. Then you add 2 more and you get 14.

Hope this helps !

3 0

3 years ago

Read 2 more answers

a $7,000 principal is invested in two accounts, one earning 3% interest and another earning 7% interest. if the total interest f

Ksju [112]

Let the investment in first account be x and this account is earning 3% interest

So Interest on this account is 3% of x = 0.03x

So the investment in second account would be (7000-x) and this account is earning 7% interest

So the interest on this account is 7% of (7000-x) = 0.07*(7000-x)

Given that total interest for the year is $262

So we have

0.03x + 0.07(7000-x) = 262

0.03x - 0.07x + 0.07*7000 = 262

-0,04x + 490 = 262

-0.04x = 262-490

-0.04x = - 228

0.04x = 228x

x = 228/0.04

x = 5700

Therefore Investment in one account s 5700 and the investment in other account is 7000-5700 = 1300

3 0

3 years ago

Read 2 more answers

Evaluate the following integral (Calculus 2) Please show step by step explanation!

Nuetrik [128]

Answer:

$4\ln \left| \dfrac{1}{3}\sqrt{9+(\ln x)^2} + \dfrac{1}{3}\ln x \right|+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{4}{x\sqrt{9+(\ln(x))^2}}\:\:\text{d}x$

Rewrite 9 as 3²:

$\implies \displaystyle \int \dfrac{4}{x\sqrt{3^2+(\ln(x))^2}}\:\:\text{d}x$

Integration by substitution

$\boxed{\textsf{For }\sqrt{a^2+x^2} \textsf{ use the substitution }x=a \tan\theta}$

$\textsf{Let } \ln x=3 \tan \theta$

$\begin{aligned}\implies \sqrt{3^2+(\ln x)^2} & =\sqrt{3^2+(3 \tan\theta)^2}\\ & = \sqrt{9+9\tan^2 \theta}\\ & = \sqrt{9(1+\tan^2 \theta)}\\ & = \sqrt{9\sec^2 \theta}\\ & = 3 \sec\theta\end{aligned}$

Find the derivative of ln x and rewrite it so that dx is on its own:

$\implies \ln x=3 \tan \theta$

$\implies \dfrac{1}{x}\dfrac{\text{d}x}{\text{d}\theta}=3 \sec^2\theta$

$\implies \text{d}x=3x \sec^2\theta\:\:\text{d}\theta$

Substitute everything into the original integral:

$\begin{aligned} \implies \displaystyle \int \dfrac{4}{x\sqrt{9+(\ln(x))^2}}\:\:\text{d}x & = \int \dfrac{4}{3x \sec \theta} \cdot 3x \sec^2\theta\:\:\text{d}\theta\\\\ & = \int 4 \sec \theta \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle 4 \int \sec \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{7 cm}\underline{Integrating $\sec kx$}\\\\$\displaystyle \int \sec kx\:\text{d}x=\dfrac{1}{k} \ln \left| \sec kx + \tan kx \right|\:\:(+\text{C})$\end{minipage}}$