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2 years ago

WILL GIVE BRAINLIEST! Find the measure of ∠a. (Let b = 66°, and let c = 172°.) m∠a = °

Mathematics

1 answer:

storchak [24]2 years ago

4 0

Answer: a = 122*

The three angles added together form a circle or 360*.

Set a + b + c = 360 and solve for a

a + 66 + 172 = 360

Subtract 238 from both sides to get a

a = 122*

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Can all quadrilaterals be inscribed in a circle? Explain

-BARSIC- [3]

Depends of the size of the quadrilateral in relation to the circle, but otherwise, yes

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3 years ago

Pls answer this i need help

Vitek1552 [10]

Step-by-step explanation:

your don't see what went wrong ?

Tyrell worked on

16 - 2(4 + 3x)

what do brackets mean ? they mean that this operation has to be done before everything else in the expression.

as this contains a variable, we cannot fully calculate it, true, but we need to keep this in mind and always treat the content of the brackets as one package.

so, whatever I do from the outside with one part of that package, I have to do also with all the other parts of that package.

so, the multiplication with -2 has to happen with both : 4 and 3x. not just with 4.

therefore, the correct simplification looks like

16 - 8 - 6x

8 - 6x or -6x + 8

Amelia multiplied correctly, but then made a mistake summing things up

10x - 3(4x + 1)

10x - 12x - 3

10x - 12x = -2x

I can't mix the pure constant -3 into the factors of x. that would be like the famous mixing of apples and oranges.

so, the result is

-2x - 3

6 0

3 years ago

The student-to-faculty ratio at a small college is 17 to 3. The total number of students and faculty is 740. How many faculty me

ryzh [129]

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3 years ago

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

5) -8x – 6 y = 13<br> 4x+3y=-9

Pepsi [2]

Answer:

no solution exists

Step-by-step explanation:

4 0

3 years ago