All categories

3 years ago

Can somebody actually help me?

Mathematics

2 answers:

Svetach [21]3 years ago

7 0

The answer is letter B.

puteri [66]3 years ago

5 0

Answer:

Step-by-step explanation:

u can count the amount of steps the triangle has been shifted. it's 3 up and 5 left. leaving the red triangle I it's new position

You might be interested in

What value of x makes the equation true 4x = 32?

pochemuha

Answer:

$x=8$

Step-by-step explanation:

$4x=32$

$4\times x =32$

Divide both sides by 4:

$\frac{4\times x}{4}=\frac{32}{4}$

$x=8$

5 0

4 years ago

Read 2 more answers

Penelope is going to invest $23,000 and leave it in an account for 11 years. Assuming the interest is compounded continuously, w

madreJ [45]

Answer:

5.3%

Step-by-step explanation: answer fo

r delta math

6 0

3 years ago

Complete the remainder of the<br> table for the given function rule:

Alexus [3.1K]

-1, -8
0, -9
1, -10
2, -11
Hope that helps!

7 0

3 years ago

How to solve:<br> 8(2)-4<br>________<br>8÷4 ?

MissTica

So the first thing you do is dove the equations. Let's do the numerator equation. 8(2)-4 is simply saying 8•2-4 and i don't know if u learned this in class yet but you do multiplication and division before addition and subtraction so 8•2=16-4=12 so now 12 is our numerator. Now for the denominator, 8/4=2 so 2 is our denominator. We have 12/2 but it can be simplified to 6 because 6 goes into 12 twice and u cans check this by doing 6•2=12
Hope this helps m8 :))

6 0

4 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