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2 years ago

In XYZ, XY=XZ Find the length of XY of XYZ if XY=2a, YZ=3a+1, andXZ=5a-12

Mathematics

1 answer:

lorasvet [3.4K]2 years ago

5 0

Answer: XY = 8

Step-by-step explanation:

The diagram of the triangle XYZ is shown in the attached photo.

XY=XZ. This means that two sides of the triangle are equal. Two angles are also equal. It means that the triangle XYZ is an isosceles triangle.

Since XY=XZ, then

2a = 5a - 12

Subtracting 2a from both sides of the equation,

2a - 5a = 5a - 12 - 5a

-3a = -12

a = -12/-3

a = 4

Therefore, the length of XY is 2a = 2×4 = 8

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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

I will give brainliest answer

stich3 [128]

Answer:

Left 3, down 2

Step-by-step explanation:

(6,4) --> (3,2)

-3, -2

Left 3, down 2

6 0

3 years ago

Drag a number to each box to express 0.000000015 in scientific notation.

loris [4]

In a scientific notation, it would be: 1.5 × 10⁻⁸

In short, Your Answer would be 1.5 × 10⁻⁸

Hope this helps!

8 0

3 years ago

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lbvjy [14]

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All the points on a line that are between two given points form a line segment

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3 years ago

Stu is x years old. Paul is 4 years older than Stu. Six years ago Paul was how many years old?

Luden [163]

This is a badly worded problem

4 0

3 years ago