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Feliz [49]
3 years ago
6

The perimeter of a rectangular deck is 4 times its width. If the deck's perimeter is 30 feet, what is the deck's area?

Mathematics
1 answer:
RSB [31]3 years ago
4 0

Answer:

i think it is 120

Step-by-step explanation:

4 multiplied by 30 is 120

You might be interested in
The cosine of 23° is equivalent to the sine of what angle
Archy [21]

Answer:

So 67 degrees is one value that we can take the sine of such that is equal to cos(23 degrees).

(There are more values since we can go around the circle from 67 degrees numerous times.)

Step-by-step explanation:

You can use a co-function identity.

The co-function of sine is cosine just like the co-function of cosine is sine.

Notice that cosine is co-(sine).

Anyways co-functions have this identity:

\cos(90^\circ-x)=\sin(x)

or

\sin(90^\circ-x)=\cos(x)

You can prove those drawing a right triangle.

I drew a triangle in my picture just so I can have something to reference proving both of the identities I just wrote:

The sum of the angles is 180.

So 90+x+(missing angle)=180.

Let's solve for the missing angle.

Subtract 90 on both sides:

x+(missing angle)=90

Subtract x on both sides:

(missing angle)=90-x.

So the missing angle has measurement (90-x).

So cos(90-x)=a/c

and sin(x)=a/c.

Since cos(90-x) and sin(x) have the same value of a/c, then one can conclude that cos(90-x)=sin(x).

We can do this also for cos(x) and sin(90-x).

cos(x)=b/c

sin(90-x)=b/c

This means sin(90-x)=cos(x).

So back to the problem:

cos(23)=sin(90-23)

cos(23)=sin(67)

So 67 degrees is one value that we can take the sine of such that is equal to cos(23 degrees).

6 0
3 years ago
Weather affects flights across the nation, causing delays, cancellations, and general disruption to the NAS. You have gathered a
hammer [34]

Answer:

a.) one sample t test

b.) H0 : μ = 59.3

c.) H1 : μ > 59.3

d.) μ = 59.3 ; σ = 39.84

e.) xbar = 79.4 ; s = 61.36

Test statistic = 3.16

Step-by-step explanation:

Given the sample data:

49.00 49.00 49.00 49.00 49.00 63.00 63.00 63.00 63.00 63.00 199.00 199.00 199.00 199.00 199.00 38.00 38.00 38.00 38.00 38.00 48.00 48.00 48.00 48.00 48.00 49.00 63.00 199.00 38.00 48.00

Sample size, n = 30

Using calculator :

xbar from the data above = 79.4

Standard deviation = 61.359

H0 : μ = 59.3

H1 : μ > 59.3

Test statistic :

(Xbar - μ) ÷ (σ/sqrt(n)

σ = 34.83

(79.4 - 59.3) ÷ (34.83/sqrt(30))

20.1 ÷ 6.359

Test statistic = 3.16

4 0
3 years ago
Plz i need help :ppppppp
miv72 [106K]

Answer: D

Step-by-step explanation:

Just do 7 x 88 LOL

8 0
3 years ago
If susan, janice, and melvin take 2.0 hours to walk to their school at a rate of 1.0 m/s, how far is their school from their hou
RideAnS [48]

Answer:

7.2 km

Step-by-step explanation:

Given that:

Hours taken = 2 hours

Walk rate = 1 m/s

Distance = Speed * time

Time taken in seconds = 2 * 60 * 60 = 7200 s

Hence,

Distance = 1 m/s * 7200s

Distance = 7200 meters

Distance in km = 7200 / 1000 = 7.2 km

3 0
3 years ago
You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling
BARSIC [14]

Answer:

  (a) 15 m

  (b) ground speed: 26.14 m/s (94.1 km/h); relative speed 12 m/s

Step-by-step explanation:

We can choose the frame of reference to be the trailing car. We can start counting time from the point the lead car begins deceleration. Then its position (in meters) relative to the trailing car is ...

  d(t) = 25 - 5/2t² . . . . . . where 25 m is the initial distance and -5 is the acceleration in m/s².

(a) Then the distance between cars at t=2 is ...

  d(2) = 25 - (5/2)·4 = 15 . . . . meters

__

(b) The <em>relative velocity</em> at 2.4 seconds is ...

  d'(t) = -5t

  d'(2.4) = -5(2.4) = -12.0 . . . m/s

The closure speed with the police car is 12 m/s.

__

We need to do a little more work to find the ground speed of the trailing car when the cars bump.

The distance between the cars at t=2.4 seconds is ...

  d(2.4) = 25 -(5/2)2.4² = 10.6 . . . . meters

At the closure speed of 12 m/s, it will take an additional ...

  10.6 m/(12 m/s) = 0.8833... s = 53/60 s

to close the gap. The speed of the trailing car at that point in time will be the original speed less the deceleration for 0.8833 seconds:

  (110 km/h)·(1/3.6 (m/s)/(km/h))-(5 m/s²)(53/60 s)

     = 26 5/36 m/s = 94.1 km/h

_____

<em>Comment on following distance</em>

To avoid collision, the trailing car must be trailing by at least the distance it covers in 2.4 seconds, about 73 1/3 meters.

6 0
3 years ago
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