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4 years ago

How do you solve You earn 15n 15n dollars for mowing n lawns. How much do you earn for mowing one lawn? Seven lawns?

Mathematics

1 answer:

Rudik [331]4 years ago

8 0

Answer:

$15 for a lawn, $105 for seven lawns.

Step-by-step explanation:

The question is about the direct ratio problem. Logically, you will earn more if you mow more lawn.

The scheme for this ratio is given below:

15 n_________n lawns

x_________ 1 lawn

$x=\frac{15n*1}{n} = 15$

You will earn 15 dollars for mowing a lawn

When you mow 7 lawns, the scheme will be in this manner:

15 n__________n lawns

x__________7 lawns

$x=\frac{15n*7}{n}=105$

You will earn 105 dollars

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Greatest common factor for 3,27, and 33

Georgia [21]

Answer:

it's 3.

Step-by-step explanation:

Here's why........3 only has 2 factors 1 and it's self and 3 times 9 is 27 and 3 times 11 is 33 they all have 3 as a factor.

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3 years ago

Find the average rate of change of the function over the given interval.

Minchanka [31]

Answer:

Average rate of change = 20.2

Rate of change at the left endpoint : f' (5) = 4t = 20

Rate of change at the right endpoint : f' (5.1) = 4*5.1 = 20.4

Step-by-step explanation:

The average rate of change of the function

F(t) = 2t^2 - 1 , [ 5,5.1 ]

solution

$\frac{f(5.1)-f(5)}{5.1 - 5}$ = $\frac{2*(5.1)^2-1-(2*5^2-1)}{0.1}$ = [ 2 ( 26.01 - 25 ) / 0.1 ]

= 2.02 / 0.1 = 20.2

Rate of change at the left endpoint : f' (5) = 4t = 20

Rate of change at the right endpoint : f' (5.1) = 4*5.1 = 20.4

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4 years ago

Need help look at picture. Will Mark Brainliest and only answer if your 100% sure.

lina2011 [118]

Answer: 207

Step-by-step explanation:

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3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago

A student’s first step to solving the equation 5(x – 2) = 10 was to simplify it to 5x – 2 = 10. Did the student make a mistake?

Dafna11 [192]

Yes, he forgot to distribute all the terms.

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4 years ago

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