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3 years ago

−5y+8x=−18 5y+2x=58

Mathematics

2 answers:

JulsSmile [24]3 years ago

8 0

Cancel -5y and 5x
leaving you with
8x=-18
2x=58
add
10x = 40
x= 4
substitute to find y
5y+2x=58
5y+8=58
5y=50
y=10

Mrrafil [7]3 years ago

8 0

Answer: x = 4,y=10

Step-by-step explanation:

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I need help with this plz

Ksju [112]

Answer:

The correct answer is D. 10.

Step-by-step explanation:

If Brandy earns 2 stars for every goal, and she scored 8 goals, then we need to multiply 8 by 2.

8 x 2 = 16

She looses 1/2 a star for every miss, and she missed the goal 12 times. Therefore, we need to multiply 1/2 by 12.

12 x 1/2 = 6

Therefore, Brandy lost 6 stars.

Now, we need to subtract the amount of stars Brandy earned by the amount that she lost.

16 - 6 = 10

Therefore, the correct answer is D. 10.

Hope this helps! :D

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3 years ago

What is the area of the circle of radius 9

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3 years ago

Read 2 more answers

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

What is the value of (x) = -3.25x + 22.41 at x = -4.2?

victus00 [196]

Don’t mind this comment just need points but good luck

6 0

3 years ago

Why does dividing by 10 shift each digit one place to the right?Clear

Sedbober [7]

Answer:

D. Each digit represents 10 times its original value.

Step-by-step explanation:

8 0

3 years ago

−5y+8x=−18 5y+2x=58 ​

−5y+8x=−18 5y+2x=58