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valentinak56 [21]

3 years ago

13

the diagram shows a triangle 3x-1,3x and 4x-1. The perimeter of the triangle is 70 cm. The area of the triangle is a^2 cm. What

is the value of a?

1 answer:

pychu [463]3 years ago

8 0

Answer:

4x

Step-by-step explanation:

45

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What is the solution to <br> 5x+9-3x=18+15<br><br> A.x=8<br> B.x=12<br> C.x=21<br> D.x=3

Vlad [161]

Your answer will be B) x=12

1)5x-3x
2) move 9 to the other side (change signs)
3)18+15-9
4) 2x= 24 (divide 2 from both sides
5)x=12

hope this helps

7 0

3 years ago

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Instructions: Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Morgarella [4.7K]

Answer:

y = 3x - 13................slope= 3, point= (5,2)

y = 3x + 11.................slope= 3, point= (2,17)

3y = x + 11................slope= 1/3, point= (-5,2)

3y = x - 13................slope= 1/3, point= (19,2)

Step-by-step explanation:

To match each equation to its slope and point, remember that an equation in y = mx + b has slope m. The point must also satisfy or make true the equation when substituted for (x,y).

y=3x-13 has slope 3. The point (2,17) make it false. But (5,2) makes it true.

17 = 3(2) - 13 2 = 3(5) - 13

17 = 6-13 2 = 15 - 13

17 = - 7 False 2 = 2 True

y=3x+11 has slope 3. This means (2,17) must be true.

3y=x+11 can be converted to y = 1/3x + 11/3. It has slope 1/3. Test each point.

2= 1/3(19) + 11/3 -2 = 1/3(17) + 11/3 2 = 1/3(-5) + 11/3

2 = 19/3 + 11/3 -2 = 17/3 + 11/3 2 = -5/3 + 11/3

False False 2 = 6/3 True

3y=x-13 can be converted to y = 1/3x - 13/3. It has slope 1/3. Test each remaining point.

2= 1/3(19) - 13/3 -2 = 1/3(17) - 13/3

2 = 19/3 - 13/3 -2 = 17/3 - 13/3

2 = 6/3 -2 = 4/3

True False

7 0

3 years ago

Read 2 more answers

Simplify 7 - (-4) + 3(-6)

jeka94

Answer:

-7

Step-by-step explanation:

7 - (-4)= 11

+

3(-6)= -18

=-7

8 0

3 years ago

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When would u need an interpreter? What would that person do?

seropon [69]

If you speak a different language this person can interpret your speaking into the language understandable by the other party

6 0

3 years ago

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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