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3 years ago

Find the value of b. Then find the angle measures of the pentagon.

Mathematics

1 answer:

gizmo_the_mogwai [7]3 years ago

8 0

B + 3/2*b + b + 45 + 2b - 90 + 90 = 540
Or, 4b + 3b/2 + 45 = 540
Or, (8b+3b)/2 = 495
Or, 11b = 990
Or, b = 90
Now,
b+45 = 135
3/2*b = 135
2b - 90 = 90

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Which of the following is equal to 2,952 ÷ 24?

attashe74 [19]

Answer:

Step-by-step explanation:

After doing long division we then know that 2,952 ÷24 = 123

We 1st follow pemdas knowing this we solve the equations in parenthesis 1st

(2,400 ÷ 24) + (480 ÷ 24) + (72 ÷ 24)

2,400 ÷ 24 = 100

480 ÷ 24 = 20

72 ÷ 24 = 3

We can then rewrite the equation as

100 + 20 + 3 We then solve left to right

100 + 20 = 120

120 + 3 = 123

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2 years ago

PLEASSE HEP MEEEEEEEE

DENIUS [597]

Answer:

s = 70

Step-by-step explanation:

$\frac{s}{10} = 7$

$10 \times \frac{s}{10} = 10 \times 7$

$s = 70$

<h2>Insta: 25k_kem</h2>

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2 years ago

Which new angle is created by extending one side of a triangle

8090 [49]

Answer: The correct option is (C). an exterior angle.

Step-by-step explanation: We are to select the name of the new angle that is created by extending one side of a triangle.

Let ABC be a triangle as shown in the attached figure.

∠ABC, ∠ACB and ∠BAC are three interior angles of the triangle ABC.

Let us extend the side BC towards C to point D. The newly created angle is ∠ACD.

∠ACD is called the exterior angle of ΔABC.

Therefore, the newly created angle is an exterior angle.

Thus, option (C) is correct.

7 0

3 years ago

Read 2 more answers

The mass of an electron is approximately 9 × 10-28 grams, while the mass of a neutron is approximately 2 × 10-24 grams. Which of

aalyn [17]

Answer: the mass of a neutron is approximately 2,000 times the mass of an electron

Step-by-step explanation:

- the easiest way to solve this (in my opinion) is to simply divide the mass of a neutron by the mass of an electron

- $2 x10^{-24} / (9 x10^{-28} )$

= $(2/9) x10^{-24--28}$

= $(2/9)x10^{-24+28}$

≈ $0.2222x10^{28-24}$

≈ $0.2222x10^{4}$

≈ which is approximately 2222

- so 2222 is approximately 2000 times

- therefore, the mass of a neutron is approximately 2,000 times the mass of an electron

hope this helps :)

4 0

2 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