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WINSTONCH [101]
4 years ago
12

Two planes make a 1500 mile flight, one flying 25 miles per hour faster than the other. The quicker plane makes the trip 3 hours

faster. How long did it take the slower plane to complete the flight?
Mathematics
1 answer:
oksian1 [2.3K]4 years ago
3 0
1500/25=60
60-3=57
It took 57 hours for the slower plane to complete the flight, and it was flying at 19mph. My question is, how did a plane stay in the sky going only 25 miles per hour? Let alone even 19 miles per hour? I can pedal a bicycle faster than that!
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Which shows the graph of the solution set of 3y – 2x > –18?
nikdorinn [45]

Answer:

Graph of the inequality 3y-2x>-18 is given below.

Step-by-step explanation:

We are given the inequality, 3y-2x>-18

Now, using the 'Zero Test', which states that,

After substituting the point (0,0) in the inequality, if the result is true, then the solution region is towards the origin. If the result is false, then the solution region is away from the origin'.

So, after substituting (0,0) in 3y-2x>-18, we get,  

3\times 0-2\times 0>-18

i.e. 0 > -18, which is true.

Thus, the solution region is towards the origin.

Hence, the graph of the inequality 3y-2x>-18 is given below.

8 0
3 years ago
rhombus CDEF is shown below. If the slope of FC is 25, what must be the slope of CD in order for CDEF to be a square?
Andrew [12]

Answer:

-\dfrac{1}{25}

Step-by-step explanation:

Given the rhombus CDEF, sides FC and CD are adjacent sides.

Using coordinate geometry, to proof that a quadrilateral is a square, with regards to the slope, the slope of adjacent segments must be negative reciprocals.

In fact, by definition: Two lines with slopes m_1$ and m_2 are perpendicular if:

m_1=-\dfrac{1}{m_2}

Therefore if :

  • FC and CD are adjacent segments
  • Slope of FC=25

For the rhombus to be a square:

Slope of CD =-\dfrac{1}{25}

8 0
3 years ago
#12-10: Factor this expression by dividing by the GCF first. 4h+24
AleksAgata [21]

\huge{\boxed{4(h+6)}}

Let's start by writing out their factors.

4: <em><u>1</u></em><em>, </em><em><u>2</u></em><em>, </em><em><u>4</u></em>

24: <em><u>1</u></em><em>, </em><em><u>2</u></em><em>, 3, </em><em><u>4</u></em><em>, 6, 8, 12, 24</em>

The greatest factor that 4 and 24 have in common is 4.

Factor out 4 from both terms. \boxed{4(h+6)}

6 0
3 years ago
Find the area of the blue area if each<br> square represents 1.5 square meters.
Anni [7]

Check the picture below.

so the picture has a rectangle that is 8 units high and 12 units wide, and it has a couple of "empty" trapezoids, with a height of 5 and "bases" of 9 and 3.

now, if we just take the whole area of the rectangle and then subtract the area of those two trapezoids, what's leftover is the blue area.

\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ h=5\\ a=9\\ b=3 \end{cases}\implies \begin{array}{llll} A=\cfrac{5(9+3)}{2}\implies A=30 \end{array} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large Areas}}{\stackrel{rectangle}{(12\cdot 8)}~~ -~~\stackrel{\textit{two trapezoids}}{2(30)}}\implies 96-60\implies 36

3 0
2 years ago
PLS SOMEONE LOOK AT THE PHOTO I POSTED AND ANSWER AS SOON AS POSSIBLE PLS ITS ASKING FOR X= AND Y= NOT AN EQUATION JUST PLS ANSW
34kurt

We can solve it, but we can't actually use the interactive system they want you to use.  Let's see if we can guide you through it.

My guess is you can control the lines by dragging the two points on the line.

Let's make the green line y = 2x + 3

When x=0 we get y=3, point (0,3).   That's the y intercept, 3 up on the y axis.  Drag one green point there.

When x=2 we get y=7, point (2,7), two to the right, seven  up.  Drag the other green point there.   The green line is now y = 2x + 3.

Similarly the red line is made y = -3x + 3.    When x=0, y=3, point (0,3).  [Surprise!]   When x=2, y=-3, point (2,-3).  Drag the red points to (0,3) and (2,-3).

We already saw where they meet, at (0,3), x=0, y=3, the y intercept of both lines.

Answer: x=0, y=3

8 0
4 years ago
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