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3 years ago

The week before Jack spend 19 1/4 hours doing homework.how long did Jill spend doing homework

Mathematics

2 answers:

nlexa [21]3 years ago

5 0

Jill spent 19 1/4 hours doing homework last week.

Vsevolod [243]3 years ago

3 0

19 1/4 hours doing homework Jill did

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Mr.Davis drives 496 miles in eight hours. At this rate, how many miles does he drive in six hours?

Lynna [10]

Answer:

Step-by-step explanation:

496m=8h

Figure for 1 hour.

496/8=62

Now do

62x6=372m

Yes, you are correct.

3 0

3 years ago

A force of 3 pounds is required to hold a spring stretched 0.6 feet beyond its natural length. how much work (in foot-pounds) is

ioda

The work done (in foot-pounds) in stretching the spring from its natural length to 0.7 feet beyond its natural length is 1.23 foot-pound

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

Force (F) = 3 pounds
Extension (e) = 0.6 feet
Work done (Wd) =?

<h3>How to determine the spring constant</h3>

Force (F) = 3 pounds
Extension (e) = 0.6 feet
Spring constant (K) =?

F = Ke

Divide both sides by e

K = F/ e

K = 3 / 0.6

K = 5 pound/foot

Thus, the spring constant of the spring is 5 pound/foot

<h3>How to determine the work done</h3>

Spring constant (K) = 5 pound/foot
Extention (e) = 0.7 feet
Work done (Wd) =?

Wd = ½Ke²

Wd = ½ × 5 × 0.7²

Wd = 2.5 × 0.49

Wd = 1.23 foot-pound

Therefore, the work done in stretching the spring 0.7 feet is 1.23 foot-pound

Learn more about spring constant:

brainly.com/question/9199238

#SPJ1

5 0

2 years ago

Help plz plz plz plz plz plz plz plz

Rom4ik [11]

15 5/6 - 3 11/15

15 - 3 = 12
5/6 - 11/15 = 25/30 - 22/30 = 3/30 = 1/10

12 + 1/10 = 12 1/10 <== ur answer

8 0

3 years ago

Which of the following statements is true of hypothesis testing?

aleksklad [387]

Answer:

i personaly dont know the answer u could try copy and pasting in google or something, anyways look it up, im pretty sure youll have a answer

Step-by-step explanation:

5 0

3 years ago

Determine which function has the greatest rate of change over the interval [0, 2].

ruslelena [56]

Remember that the average rate of change of a function over an interval is the slope of the straight line connecting the end points of the interval. To find those slopes, we are going to use the slope formula: $m= \frac{y_{2}-y_{1}}{x_2-x_1}$

Rate of change of $a$ :
From the graph we can infer that the end points are (0,1) and (2,4). So lets use our slope formula to find the rate of change of $a$ :
$m= \frac{y_{2}-y_{1}}{x_2-x_1}$
$m= \frac{4-1}{2-0}$
$m= \frac{3}{2}$
$m=1.5$
The average rate of change of the function $a$ over the interval [0,2] is 1.5

Rate of change of $b$ :
Here the end points are (0,0) and (2,2)
$m= \frac{2-0}{2-0}$
$m= \frac{2}{2}$
$m=1$
The average rate of change of the function $b$ over the interval [0,2] is 1

Rate of change of $c$ :
Here the end points are (0,-1) and (2,0)
$m= \frac{0-(-1)}{2-0}$
$m= \frac{1}{2}$
$m=0.5$
The average rate of change of the function $c$ over the interval [0,2] is 0.5

Rate of change of $d$ :
Here the end points are (0,0.5) and (2,2.5)
$m= \frac{2.5-0.5}{2-0}$
$m= \frac{2}{2}$
$m=1$
The average rate of change of the function $d$ over the interval [0,2] is 1

We can conclude that the <span>function that has the greatest rate of change over the interval [0, 2] is the function a.</span>

4 0

4 years ago