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deff fn [24]
3 years ago
11

A jet fighter is capable of reaching speeds up to 1.5 x 10^3 miles per hour. A typical passenger plane can reach speeds of appro

ximately
5.0 x 10^2 miles per hour. About how many times faster can the jet fighter travel per hour than the typical passenger plane?
Mathematics
1 answer:
Volgvan3 years ago
5 0

Answer:

3 times faster

Step-by-step explanation:

\frac{1.5(10)^{3} }{5.0(10)^{2}} = \frac{1.5(10)}{5} = \frac{15}{5}  = 3 times faster

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A rational number that is not an integer can be expressed in decimal form with a non-zero digit after the decimal
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Which step is The correct answer?
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Step 1 is correct because it is easy method to do math equations
6 0
2 years ago
Please answer correctly !!!!!!!!!!!!!! Will Mark Brianliest !!!!!!!!!!!!!!!!!!!
alexgriva [62]

Answer:

x = 30

Step-by-step explanation:

well from the theorem we have

\frac{15}{3}=\frac{x}{6}

yes i know you could say that the right way is

\frac{3}{15}=\frac{6}{x}

well if you notice they are the same only that in my way the x is in the numerator which means it will be far easier to know it's value :)

so

\frac{15}{3}=\frac{x}{6}\\\\5=\frac{x}{6}\\\\6[5]=6[\frac{x}{6}]\\\\30=x

8 0
3 years ago
Eric works for an airline, and he needs to calculate the weight of passengers and luggage before takeoff. The number of
igomit [66]

Answer:

B

Step-by-step explanation:

To complete the question, here are the answer choices:

<em>A)  =A1*A2*A3*A4 </em>

<em>B)  =A1*A2+A3+A1*A4 </em>

<em>C)  =A1*(A2+A3+A1)*A4 </em>

<em>D)  =A1*A2+(A3+A1)*A4</em>

<em />

We first need to multiply A1 and A2, this will give weight of passengers.

To get weight of luggage, we multiply A1 and A4.

We also need the checked weight to add to that, which is in A3. So then we add up A3 with those 2.

So we will get

A1*A2 + A1*A4 + A3

This is given in a different order in Option B. Hence, option B is right.

4 0
3 years ago
Read 2 more answers
A line m is perpendicular to an angle bisector of ∠A. The sides of ∠A intersect this line m at points M and N. Prove that △AMN i
Ray Of Light [21]

Answer:


Step-by-step explanation:

<em><u>Given</u></em><u>:</u>       A line m is perpendicular to the angle bisector of ∠A. We call this  

                 intersecting point as D. Hence, in figure ∠ADM=∠ADN =90°.

                 AD is angle bisector of ∠A. Hence, ∠MAD=∠NAD.

<u><em>To Prove</em></u>:   <em><u>ΔAMN is an isosceles triangle. i.e any two sides in ΔAMN are</u></em>

<em>                    </em><em><u>equal. </u></em>

<em><u>Solution</u></em>:  Now, In ΔADM and ΔADN

                 ∠MAD=∠NAD     ...(1) (∵Given)

                  AD=AD                ...(2) (∵common side)

                  ∠ADM=∠ADN     ...(3) (∵Given)

                  <u><em> Hence, from equation (1),(2),(3) ΔADM ≅ ΔADN</em></u>

                                                         ( ∵ ASA  congruence rule)

                  ⇒<u><em> AM=AN</em></u>

                  Now, In Δ AMN

                 AM=AN (∵ Proved)

                  Hence, ΔAMN is an isosceles  triangle.


7 0
3 years ago
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