Help me this is math pls!

Find the value of y.

Natali [406]

Answer:

y=7

Step-by-step explanation:

The two given angels are congruent so they must be equal in measure.

Set them up equal and find y.

8y = 5y + 21, subtract 5y from both sides

8y -5y = 21, combine like terms

3y = 21, divide both sides by 3

y = 7

7 0

2 years ago

Can any one help me on this? Please thank you

nignag [31]

The area of a triangle is base length x height then divide all by 2,
so 12 x 10 /2
=60 ft²

5 0

3 years ago

If sum of first 6 digits of AP is 36 and that of the first 16 terms is 255,then find the sum of first ten terms.

Cloud [144]

Answer:

100

Step-by-step explanation:

We have the sum of first n terms of an AP,

Sn = n/2 [2a+(n−1)d]

Given,

36= 6/2 [2a+(6−1)d]

12=2a+5d ---------(1)

256= 16/2 [2a+(16−1)d]

32=2a+15d ---------(2)

Subtracting, (1) from (2)

32−12=2a+15d−(2a+5d)

20=10d ⟹d=2

Substituting for d in (1),

12=2a+5(2)=2(a+5)

6=a+5 ⟹a=1

∴ The sum of first 10 terms of an AP,

S10 = 10/2 [2(1)+(10−1)2]

S10 =5[2+18]

S10 =100

This is the sum of the first 10 terms.

Hope it will help.

4 0

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

3. Find the length of the missing side given the information below.

Lubov Fominskaja [6]

<h3>Answer: 12.5 (choice C)</h3>

=================================================

We apply the pythagorean theorem to find this answer.

a = 11 and b = 6 are the given legs

c = unknown hypotenuse

So,

a^2+b^2 = c^2

c = sqrt( a^2+b^2 )

c = sqrt( 11^2 + 6^2 )

c = sqrt( 121 + 36 )

c = sqrt( 157 )

c = 12.52996 approximately

c = 12.5

Side note: once you replace 'a' and b with 11 and 6, you can compute everything with a calculator in one single step more or less. The steps above are shown if you wanted to find the exact value sqrt(157).

7 0

3 years ago