1.What is the slope of this graph?

Greta added 3/6 of a cup of sugar to her cake mix, then sprinkled

bazaltina [42]

3/6 of a cup, 1/6 times 3/1 equals 3/6 or 1/2 :)

8 0

3 years ago

Which equation is a direct variation? PLEASE HELP

lyudmila [28]

It would be your first choice, y= 3/5x i believe

6 0

3 years ago

A family pays $21.99 each month for its long distance phone service. This is 80% of the original price of the phone service. Wha

gladu [14]

Answer:

The original price of the phone service is $27.49.

Step-by-step explanation:

With the information provided, you can use a rule of three to find the original price that would represent 100% given that $21.99 represents 80% of that price:

21.99 → 80%

x ← 100%

x=(21.99*100)/80

x=27.49

According to this, the answer is that the original price of the phone service is $27.49.

8 0

3 years ago

A yo-yo is moving up and down a string so that its velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. The initial pos

jeka57 [31]

Part A - The average value of v(t) over the interval (0, π/2) is 6/π

Part B - The displacement of the yo-yo from time t = 0 to time t = π is 0 m

Part C - The total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

<h3>Part A: Find the average value of v(t) on the interval (0, π/2)</h3>

The average value of a function f(t) over the interval (a,b) is

$f(t)_{avg} = \frac{1}{b - a} \int\limits^b_a {f(t)} \, dx$

So, since velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. Its average value over the interval (0, π/2) is given by

$v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {v(t)} \, dt$

Since v(t) = 3cost, we have

$v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {3cos(t)} \, dt\\= \frac{3}{\frac{\pi }{2}} \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= \frac{6}{{\pi}} [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= \frac{6}{{\pi}} [{sin(\frac{\pi }{2})} - sin0]\\ = \frac{6}{{\pi}} [1 - 0]\\ = \frac{6}{{\pi}} [1]\\ = \frac{6}{{\pi}}$

So, the average value of v(t) over the interval (0, π/2) is 6/π

<h3>Part B: What is the displacement of the yo-yo from time t = 0 to time t = π?</h3>

To find the displacement of the yo-yo, we need to find its position.

So, its position x = ∫v(t)dt

= ∫3cos(t)dt

= 3∫cos(t)dt

= 3sint + C

Given that at t = 0, x = 3. so

x = 3sint + C

3 = 3sin0 + C

3 = 0 + C

C = 3

So, x(t) = 3sint + 3

So, its displacement from time t = 0 to time t = π is

Δx = x(π) - x(0)

= 3sinπ + 3 - (3sin0 + 3)

= 3 × 0 + 3 - 0 - 3

= 0 + 3 - 3

= 0 + 0

= 0 m

So, the displacement of the yo-yo from time t = 0 to time t = π is 0 m

<h3>Part C: Find the total distance the yo-yo travels from time t = 0 to time t = π. (10 points)</h3>

The total distance the yo-yo travels from time t = 0 to time t = π is given by

$x(t) = \int\limits^{\pi}_0 {v(t)} \, dt\\= \int\limits^{\pi }_0 {3cos(t)} \, dt\\= 3 \int\limits^{\pi }_0 {cos(t)} \, dt\\ = 3 \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt + 3\int\limits^{\pi }_{\frac{\pi }{2}} {cos(t)} \, dt\\= 3 \times 2\int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= 6 [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= 6[{sin\frac{\pi }{2} - sin0]\\\\= 6[1 - 0]\\= 6(1)\\= 6$

So, the total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

Learn more about average value of a function here:

brainly.com/question/15870615

#SPJ1

4 0

1 year ago

Suppose that the functions p and q are defined as follows.

Brut [27]

<h2>Answer:</h2>

Answer:

(r o q)(-1) = 20

(q o r)(-1) = -11

Step-by-step explanation:

Given

q(x) = -2x + 1q(x)=−2x+1

r(x) = 2x^2 + 2r(x)=2x2+2

Solving (a): (r o q)(-1)

In function:

(r o q)(x) = r(q(x))

So, first we calculate q(-1)

q(x) = -2x + 1q(x)=−2x+1

q(-1) = -2(-1) + 1q(−1)=−2(−1)+1

q(-1) = 2 + 1q(−1)=2+1

q(-1) = 3q(−1)=3

Next, we calculate r(q(-1))

Substitute 3 for q(-1)in r(q(-1))

r(q(-1)) = r(3)

This gives:

r(x) = 2x^2 + 2r(x)=2x2+2

r(3) = 2(3)^2 + 2r(3)=2(3)2+2

r(-1) = 2*9 + 2r(−1)=2∗9+2

r(-1) = 20r(−1)=20

Hence:

(r o q)(-1) = 20

Solving (b): (q o r)(-1)

So, first we calculate r(-1)

r(x) = 2x^2 + 2r(x)=2x2+2

r(-1) = 2(-1)^2 + 2r(−1)=2(−1)2+2

r(-1) = 2*1 + 2r(−1)=2∗1+2

\begin{gathered}r(-1) = 6\\\end{gathered}r(−1)=6

Next, we calculate r(q(-1))

Substitute 6 for r(-1)in q(r(-1))

q(r(-1)) = q(6)

q(x) = -2x + 1q(x)=−2x+1

q(6) = -2(6) + 1q(6)=−2(6)+1

q(6) =- 12 + 1q(6)=−12+1

q(6) = -11q(6)=−11

Hence:

(q o r)(-1) = -11

8 0

3 years ago