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GenaCL600 [577]
3 years ago
5

The variables y and x have a proportional relationship, and y = 24 when x = 16.

Mathematics
2 answers:
Goryan [66]3 years ago
7 0

Answer:

x=24 or B is the correct answer.

I took the test, and this was the correct answer. Hope this helped-!

frosja888 [35]3 years ago
3 0

Answer:

b) x = 24

Step-by-step explanation:

if y is 24 and x is 16, they make the fraction \frac{24}{16}

this can be simplified to \frac{3}{2} if you divide both sides of the fraction by 8.

36/3 is 12, and 12 * 2 = 24.

so, x = 24

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10 - 5 2/3 = ? not in decimal form only fraction .
Dmitrij [34]

it would equal the same mixed number. 5 2/3

3 0
3 years ago
Read 2 more answers
HELP WITH THIS QUESTION PLEASE!!
svetlana [45]

m<PQS = m<SQR

4y - 10 = 2y + 10

2y = 20

y = 10

m<PQR = m<PQS + m<SQR

= 4y - 10 + 2y + 10

= 6y

= 6•10

= 60

3 0
3 years ago
PLEASE HELP ASAP!!<br> means a lot!!
egoroff_w [7]
The correct answer is:
B) \: y + 1 =  \frac{4}{3} (x - 9)
6 0
4 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
Help please I'm stuck show and explain your work
Deffense [45]

The expressions D, E, and F represent a correct solution to the equation.

Step-by-step explanation:

Step 1:

First, we need to solve the given equation and find the value of x which satisfies the equation.

6(x+4)=20, (x+4) = \frac{20}{6}   = 3.333,

x = 3.333 - 4 = -0.666.

So the value of x for the given equation is -0.666.

Step 2:

Now we evaluate the values of the six given options to see which ones have an x value of -0.666.

A. \frac{20-4}{6}= \frac{16}{6}  = 2.666,

B. \frac{1}{6} (20-4)= \frac{1}{6} (16)=\frac{16}{6} = 2.666,

C. 20 -6 -4 = 20 - 10 = 10.

D. \frac{20}{6} -4= 3.333 - 4 = -0.666,

E. \frac{1}{6} (20-24) = \frac{1}{6} (-4)= -0.666,

F. (20-24)\frac{1}{6} = (-4)\frac{1}{6} = -0.666.

So the options D, E, and F represent a correct solution to the given equation.

7 0
4 years ago
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