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3 years ago

Please help me solve this

Mathematics

1 answer:

N76 [4]3 years ago

5 0

Answer:

area of the square=18×18

=324 cm²

area of the circle=πr²

=3.142×9² (radius=18cm/2=9cm). =254.502 cm²

area of the shaded region=324-254.502

=69.498 cm²

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PLEASE HELP ASAP 25 PTS + BRAINLIEST TO RIGHR/BEST ANSWER. sorry for the screen

damaskus [11]

Answer:

The answer is A I just looked it up on mathaway and it is never wrong

Step-by-step explanation:

Standard form always has the largest coefficients first

7 0

3 years ago

Read 2 more answers

Suppose that 600 yards of fencing material are available to fence in

postnew [5]

Answer:

Length = 150 yards

Width = 100 yards

Step-by-step explanation:

We want 600 yards of fencing that will result in the largest 2 fenced corrals, sharing a common border.

It will take the shape of a rectangle, with a dividing fence down the center.

Let W and L, Width and Length of the larger enclosure.

See attachment.

W= Area of the larger enclosure.

The perimeter is 2W + 2L.

The dividing fence is 1W

We know that we only have 600 yards of fence, so:

2W + 2L + 1W = 600 yards

Area = W x L

---

3W + 2L = 600 (yards)

2L = 600 -3W

L = (600-3W)/2

L = 300 -(3/2)W

---

Use this expression in the Area calculation:

Area = W x L

Area = W x (300 -(3/2)W)

Area = 300W -(3/2)W^2)

To find the maximum area, take the first derivative and set to zero to find the value of W that results in the greatest area.

Area' = 300 -2(3/2)W)

0 = 300 - 3W

3W = 300

W = 100 yards

Since 3W + 2L = 600

L = (600 - 3W)/2

L = (600 - 3(100))/2

L = 150 yards

Area = 150*100 = 15,000 yards^2

8 0

3 years ago

Bob parked his car at the local garage. If his car stayed for one hour, he would have been charged $6.00. If he stayed for 4 hou

Nitella [24]

Answer:

Initial charge is $2.5

Hourly charge is $3.5/hr

Equation is y=3.5x+2.5

Step-by-step explanation:

Getting slope

Get fixed charge

Get equation

The slope can be found as follows. For 1 hour, charges is $6 so making time in x axis while money charged in y axis, the first coordinate is (1, 6) while for 4 hours with charge of $16.5 the coordinate will be (4, 16.5)

The slope will be given by dividing change in y axis to change in x axis. This will be (16.5-6)/(4-1)=3.5

To get the fixed charge

Using the equation y=mx+c where m is slope while c is fixed charge, x is time and y is total cost then for 1 hour

6=3.5(1)+c

C=6-3.5=2.5

Therefore, generally the equation will be y=3.5x+2.5

Initial charge is $2.5

Hourly charge is $3.5/hr

Equation is y=3.5x+2.5

6 0

3 years ago

A rocket is shot straight up into the air with an initial velocity of 500 ft per second and from a height of 20 feet above the g

iragen [17]

The height of the rocket above the ground after $t$ seconds is given by the equation : $H= -16t^2+Vt+h$ , where $V$ is the initial velocity and $h$ is the initial height.

Given that, $V= 500 ft/second$ and $h= 20 ft$

So, the equation will become: $H= -16t^2 +500t+20$

A) For finding the height of the rocket 3 seconds after the launch, we will plug $t=3$ into the above equation. So....

$H= -16(3)^2+500(3)+20\\ \\ H= -16(9)+1500+20\\ \\ H= -144+1500+20=1376$

So, the height of the rocket 3 seconds after the launch is 1376 feet.

B) When the rocket at a height of 400 feet, then we will plug $H= 400$

$400=-16t^2+500t+20\\ \\ 16t^2-500t-20+400=0\\ \\ 16t^2-500t+380=0\\ \\ 4(4t^2-125t+95)=0\\ \\ 4t^2-125t+95=0$

Using quadratic formula, we will get......

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(95)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{14105}}{8}\\ \\ t= 30.4705... \\ and \\ t= 0.7794...$

So, after 0.7794...seconds and 30.4705...seconds the rocket is at a height of 400 feet above the ground.

C) The time duration that the rocket remains in the air means we need to find the time taken by the rocket to reach the ground. When it reaches the ground, then $H=0$ . So.....