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2 years ago

Which table(s), if any, represents a proportional relationship? Explain.

Mathematics

1 answer:

Wewaii [24]2 years ago

4 0

Answer:

The answer is Table 2.

Step-by-step explanation:

The answer is Table 2 because it has a proportional relationship. How I know it is a proportional relationship is because it said

0:0

1:10

2:20

3:30

4:40

5:50

If you look at it closely then you will see the 10 and 1's time table.

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The equation for the speed/distance

asambeis [7]

Step-by-step explanation:

S = 93 log (d) + 65; S = wind speed; d = distance traveled by tornado

A) tornado travels 130 miles = d; find S

S = 93 log10(130) + 65

S = 261.6 mph rounds up to 262 mph

7 0

2 years ago

Read 2 more answers

Identify each number as prime or composite.Then list all of it's factors

azamat

Prime numbers : 3, 29
Factors of 3 : 1, 3
Factors of 29 : 1, 29
Composite numbers : 6, 15, 24
Factors of 6 : 1, 2, 3 , 6
Factors of 15 : 1, 3 , 5, 15
Factors of 24 : 1, 2, 3, 4, 6, 8, 12, 24

3 0

2 years ago

Use integration by parts to find the integrals in Exercise.<br> ∫(4x-12)e-8x dx.

stepladder [879]

Answer:

$e(2x^{2} -12x)-4x^{2}+C$

Step-by-step explanation:

We have been given an indefinite integral as $\int \left(4x-12\right)e-8x\:dx$ . We are asked to find the given integral.

Let us solve our given problem.

$\int \left(4x-12\right)e\:dx-\int 8x\:dx$

Take out constant:

$e\int \left(4x-12\right)\:dx-8\int x\:dx$

$e(\int 4x\:dx -\int 12\right\:dx)-8\int x\:dx$

$e(\frac{4x^{1+1}}{2} -12x)-8*\frac{x^{1+1}}{1+1}+C$

$e(\frac{4x^{2}}{2} -12x)-8*\frac{x^{2}}{2}+C$

$e(2x^{2} -12x)-4x^{2}+C$

Therefore, our required integral would be $e(2x^{2} -12x)-4x^{2}+C$ .

5 0

2 years ago

Surface area of 9cm, 10cm, and 15cm

AVprozaik [17]

Answer: The answer is 750 cm

8 0

2 years ago

Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt

ser-zykov [4K]

Answer:

$I(t)=\frac{1}{3}(1-e^{-30t})$

Step-by-step explanation:

We are given that

$\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}$

R=150 ohm

L=5 H

V(t)=10 V

$P=\frac{R}{L}=\frac{150}{5}=30$

$I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}$

$I(t)\times I.F=\int e^{30 t}\times 10 dt+C$

$I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C$

$I(t)=\frac{1}{3}+Ce^{-30 t}$

I(0)=0

Substitute t=0

$0=\frac{1}{3}+C$

$C=-\frac{1}{3}$

Substitute the values

$I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}$

$I(t)=\frac{1}{3}(1-e^{-30t})$

5 0

3 years ago