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2 years ago

4. A plane was flying at an altitude of 55,000 feet when it began the descent toward the airport. The airplan

Mathematics

2 answers:

marin [14]2 years ago

8 0

Answer:

Step-by-step explanation:

A Descent rate,

B 10725ft,

C 25min

Mamont248 [21]2 years ago

8 0

I think the anser is c:::::::::::::::::)

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The sum of three consecutive even integers is 258. write an equation ad solve to find the integers

slavikrds [6]

The consecutive integers are 85,86,87

Explanation:

the first number

the second number

the third number

(

)

(

)

258

−

255

8 0

3 years ago

Read 2 more answers

Simplify This Radical Expression

balandron [24]

Answer:

The answer is: -0.5198

Step-by-step explanation:

CUBERT(8) – CUBERT(16) is the cube root of 8 - cube root of 16:

CubeRT(8) - CubeRT(16) =

2 - 2.5198 = -0.5198

Hope this helps! Have an Awesome day!! :-)

5 0

2 years ago

Please I need help immediately

just olya [345]

Answer:

Step-by-step explanation:

4x+6<= -54

4x <= -60

x <= -15 [The point -15 is included, so it must be a solid dot]

----------------

5x + 7 > -18

5x > -25

x > -5 [The point -5 is NOT included, so it should be an empty dot.]

Solution C seems the best.

4 0

2 years ago

Saige's spaceship traveled 588 588588 kilometers ( km ) (km)left parenthesis, start text, k, m, end text, right parenthesis in 6

pshichka [43]

Answer:

The first ship has the same speed as Saige's ship while the second and third do not have the same speed.

Step-by-step explanation:

d = Distance traveled by Saige's spaceship = 588 km

t = Time taken = 60 seconds

Speed is given by

$s=\dfrac{d}{t}\\\Rightarrow s=\dfrac{588}{60}\\\Rightarrow s=9.8\ \text{km/s}$

The other ships

$s=\dfrac{441}{45}=9.8\ \text{km/s}$

$s=\dfrac{215}{25}=8.6\ \text{km/s}$

$s=\dfrac{649}{110}=5.9\ \text{km/s}$

So, the first ship has the same speed as Saige's ship while the second and third do not have the same speed.

8 0

2 years ago

denis-greek [22]

Answer:

Part a) $AB=15\ cm$

Part b) $AB=11.1\ cm$

Part c) $AB=3/5\ cm$

Part d) $AB=3s\ cm$

Step-by-step explanation:

see the attached figure to better understand the question

we know that

To find the length of the image after a dilation, multiply the length of the pre-image by the scale factor

Part a) we have

The scale factor is 5

The length of the pre-image is 3 cm

therefore

The length of the image AB after dilation is

$3(5)=15\ cm$

Part b) we have

The scale factor is 3.7

The length of the pre-image is 3 cm

therefore

The length of the image AB after dilation is

$3(3.7)=11.1\ cm$

Part c) we have

The scale factor is 1/5

The length of the pre-image is 3 cm

therefore

The length of the image AB after dilation is

$3(1/5)=3/5\ cm$

Part d) we have

The scale factor is s

The length of the pre-image is 3 cm

therefore

The length of the image AB after dilation is

$3(s)=3s\ cm$

7 0

3 years ago