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igor_vitrenko [27]

3 years ago

12

Olivia recorded the lengths of the football passes that her brother threw during a game. The numbers are shown in yards. 37, 40,

23, 62, 54 What is the average length of a pass thrown by Olivia's brother? Write your answer to the nearest tenth.

1 answer:

Aleonysh [2.5K]3 years ago

5 0

37+40+23+62+54=216. 216/5 (the number of passes) =43.2. That rounds down to 43 so the answer is 43 yards.

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Eight toy makers each carved 25 blocks and 15 cars how many toys did they crave in all

Stella [2.4K]

Answer:

320

Step-by-step explanation:

25 + 15 = 40 toys each

Toy makers x number of toys each = total toys carved

8 x 40 = 320 totals carved in all

6 0

3 years ago

Read 2 more answers

Which of the following is the solution to the differentiable equation dy/dx=4x/y, where Y(2)=-2

cupoosta [38]

Answer:

b is the answer

Step-by-step explanation:

7 0

3 years ago

URGENT PLS RESPOND!!!

Elan Coil [88]

Answer:

$\huge\boxed{d=\sqrt{17}}$

Step-by-step explanation:

Simply use the fact that the change in x squared + the change in y squared = the distance squared. (The Pythagorean Theorem)

1^2 + 4^2 = d^2

1 + 16 = d^2

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6 0

3 years ago

A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f

Andrews [41]

Answer:

$A = \frac{P}{r}\left( e^{rt} -1 \right)$

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

$\frac{dA}{rA+P} = dt$

Integration on both sides gives

$\int \frac{dA}{rA+P} = \int dt$

where $c$ is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

$\int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c$

Therefore,

$\frac{1}{r} \ln |rA+P| = t+c$

Multiply both sides by $r.$

$\ln |rA+P| = rt+c_1, \quad c_1 := rc$

By taking exponents, we obtain

$e^{\ln |rA+P|} = e^{rt+c_1} \implies |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}$

Isolate $A$ .

$rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}$

Since $A = 0$ when $t=0$ , we obtain an initial condition $A(0) = 0$ .

We can use it to find the numeric value of the constant $c$ .

Substituting $0$ for $A$ and $t$ in the equation gives

$0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P$

Therefore, the solution of the given differential equation is

$A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)$

4 0

3 years ago

Clara has a garden that is 7 feet wide and 4 feet long. she has 30 tomato plants to put in hte graden. each plants need 1 sqaure

Flauer [41]

garden: Area = 7 x 4 = 28

plants: Area = 1 x 1 x 30 plants = 30

30 - 28 = 2

Answer: 2 plants

7 0

3 years ago