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3 years ago

What is the answers to the picture

Mathematics

2 answers:

nalin [4]3 years ago

4 0

5.6/4 = 1.4(4)/4

x = 4

Montano1993 [528]3 years ago

3 0

Answer:

5.6 : 4 = 1.4(4) : 4

So, x = 4

Ps ':' means "divided"

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You want to order carpet for an area in your house. The room measures 12 feet by 10 feet. Remember, Area = Length × Width.

Wewaii [24]

Answer:

3630 dollars

Step-by-step explanation:

First, we need to find the area of the the room floor. To do this, we need to do 12 × 10 because 12 is the length and 10 is the width. 12 × 10 is 120. We also know that each square yard of carpet costs $30.25. We now have to multiply 120 × 30.25 because the room is 120 square feet large. 120 × 30.25 = 3630, so the carpet costs $3630.

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3 years ago

What is the circumference of a circle with a radius of 56 ?

kvasek [131]

<span>circumference =2(pi)r=(pi)d
use this formula.
2*(pi)*56=351.85</span>

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3 years ago

1 ) Tim earns $12.50 an hour cleaning houses. If he works from 8:00am to 2:00pm,

AVprozaik [17]

Answer:

$75

Step-by-step explanation:

First, we will find how many hours he worked. 8:00-12:00 is 4 hours, and 12:00-2:00 is 2 hours, so he worked a total of 6 hours.

Next, we have to multiply the hours worked by his hourly wage.

12.50*6=75

8 0

3 years ago

Read 2 more answers

What is 8.57 as a fraction or a mixed number

AURORKA [14]

8 $\frac{57}{100}$ or $\frac{857}{100}$

7 0

4 years ago

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Q1 A ball is thrown upwards with some initial speed. It goes up to a height of 19.6m and then returns. Find (a) The initial spee

lubasha [3.4K]

Answer:

(a) 19.6 ms⁻¹

(b) 2 s

(d) 4 s

Step-by-step explanation:

<u>Constant Acceleration Equations (SUVAT)</u>

$\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}$

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Acceleration due to gravity = 9.8 ms⁻².

When the ball reaches its maximum height, its velocity will momentarily be zero.

<u>Given values</u> (taking up as positive):

$s=19.6 \quad v=0 \quad a=-9.8$

$\begin{aligned}\textsf{Using} \quad v^2&=u^2+2as\\\\\textsf{Substitute the given values:}\\0^2&=u^2+2(-9.8)(19.6)\\0&=u^2-384.16\\u^2&=384.16\\u&=\sqrt{384.16}\\\implies u&=19.6\; \sf ms^{-1}\end{aligned}$

Therefore, the initial speed is 19.6 ms⁻¹.

Using the same values as for part (a):

$\begin{aligned}\textsf{Using} \quad s&=vt-\dfrac{1}{2}at^2\\\\\textsf{Substitute the given values:}\\19.6&=0(t)-\dfrac{1}{2}(-9.8)t^2\\19.6&=4.9t^2\\t^2&=\dfrac{19.6}{4.9}\\t^2&=4\\t&=\sqrt{4}\\\implies t&=2\; \sf s\end{aligned}$

Therefore, the time taken to reach the highest point is 2 seconds.

As the ball reaches its maximum height at 2 seconds, one second before this time is 1 s.

<u>Given values</u> (taking up as positive):

$u=19.6 \quad a=-9.8 \quad t=1$

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=19.6+(-9.8)(1)\\v&=19.6-9.8\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

The velocity of the ball one second before it reaches its maximum height is the <u>same</u> as the velocity one second after.

<u>Proof</u>

When the ball reaches its maximum height, its velocity is zero.

Therefore, the values for the downwards journey (from when it reaches its maximum height):

$u=0 \quad a=9.8 \quad t=1$

(acceleration is now positive as we are taking ↓ as positive).

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=0+9.8(1)\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

Therefore, the velocity of the ball one second before <u>and</u> one second after it reaches the maximum height is 9.8 ms⁻¹.

From part (a) we know that the time taken to reach the highest point is 2 seconds. Therefore, the time taken by the ball to travel from the highest point to its original position will also be 2 seconds.

Therefore, the total time taken by the ball to return to its original position after it is thrown upwards is 4 seconds.

4 0

1 year ago