PLEASE HELP!! URGANT!

G(t)=−(t−1)^2+5

garri49 [273]

Answer:

In the interval of - 2 ≤ t ≤ 4, the average rate of change is zero.

Step-by-step explanation:

The function is given to be $g(t) = - (t - 1)^{2} + 5$ ............ (1)

(i) Now, for 1 ≤ t ≤ 4, we will put t = 1 and t = 4.

So, g(1) = 5, g(4) = - 4. {From equation (1)}

Hence, g(1) ≠ g(4)

(ii) Now, for - 4 ≤ t ≤ - 3, we will put t = - 4 and t = - 3.

So, g(- 4) = - 20, g(- 3) = - 11. {From equation (1)}

Hence, g(- 4) ≠ g(- 3)

(iii) Now, for -2 ≤ t ≤ 0, we will put t = - 2 and t = 0.

So, g(- 2) = - 4, g(0) = 4. {From equation (1)}

Hence, g(- 2) ≠ g(0)

(iv) Now, for - 2 ≤ t ≤ 4, we will put t = - 2 and t = 4.

So, g(- 2) = - 4, g(4) = - 4. {From equation (1)}

Hence, g(- 2) = g(4)

Therefore, in the interval of - 2 ≤ t ≤ 4, the average rate of change is zero. (Answer)

5 0

4 years ago

22 ounces increased to 25 ounces

Zielflug [23.3K]

You subtract 25 by 22 then you will get positive 3
so the answer is 3 ounces

8 0

3 years ago

Find the value of x if GH and JH are both 13 units long. Justify your answer.

const2013 [10]

Answer:

$x=7$

Step-by-step explanation:

we have the triangle GHJ

If the segment side GH is equal to the segment JH

then

The triangle GHJ is an isosceles triangle

Remember that

An isosceles triangle has two equal interior angles and two equal sides

so

m∠G=m∠J

we have

m∠G=(7x+5)°

m∠J=(x+47)°

equate the equations

(7x+5)°=(x+47)°

Solve for x

$7x-x=47-5\\6x=42\\x=7$

4 0

3 years ago

What is the answers to the parts 2a-2d

il63 [147K]

2a) if 1 ft³ weighs 150 lb==>the TOTAL VOLUME =5,000/150 =33.334 ft³

2b) 1 ft³ 1,728 in³. So the TOTAL volume in in³ =33.334 x 1728 = 57,600 in³

c) Volume =(1/3)(πR²).H but R = H then V= (1/3)(πR³).plug V (=57600)

57,600 = 1/3 (πR³) ==> R³ = (3 x 57600) / π ==> R = 38 in

d) Area x thickness = Volume ==> Area x 2 in = 57600 in then:
Are =57600/2 & Area =28,800 in²

5 0

4 years ago

A Niffler is a magical creature who loves to collect shiny objects and store them in their pouch, much like a kangaroo. The amou

Fudgin [204]

Answer:

Probability that a Niffler can hold more than 32 pounds of shiny objects in their pouch is 0.1515.

Step-by-step explanation:

We are given that the amount a Niffler can hold in their pouch is approximately normally distributed with a mean of 25 pounds of shiny objects and a standard deviation of 6.8 pounds.

Let X = amount a Niffler can hold in their pouch

So, X ~ Normal( $\mu=25,\sigma^{2} =6.8^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 25 pounds

$\sigma$ = standard deviation = 6.8 pounds

Now, the probability that a Niffler can hold more than 32 pounds of shiny objects in their pouch is given by = P(X > 32 pounds)

P(X > 32 pounds) = P( $\frac{X-\mu}{\sigma}$ > $\frac{32-25}{6.8}$ ) = P(Z > 1.03) = 1 - P(Z $\leq$ 1.03)

= 1 - 0.8485 = 0.1515

The above probability is calculated by looking at the value of x = 1.03 in the z table which has an area of 0.8485.

Hence, the probability that a Niffler can hold more than 32 pounds of shiny objects in their pouch is 0.1515.

4 0

3 years ago