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Whitepunk [10]
3 years ago
9

Mark and his brother David are fighting over a toy airplane while playing their yard. Mark grabs the plane and takes off running

at rate of 7.5 miles per hour. One minute after Mark starts running, David starts to chase him at rate of 9mph. How long will David to reach Mark and how far will they have run?
Mathematics
1 answer:
-BARSIC- [3]3 years ago
5 0
David will have 2.57 seconds to get to mark and it’s 50 meters
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Please answer it’s 20 points for award!
guajiro [1.7K]

Answer:

The scale drawing is 20 units by 4 units

Step-by-step explanation:

20/5 = 4

100/5 = 20

6 0
3 years ago
The sum of the measures of angle M and angle R is 90°.
denis23 [38]

Answer:

x = 5

Step-by-step explanation:

We have: The sum of the measures of angle M and angle R is 90°

M + R = 90°    

M = (5x + 10)°

Plug M into (1)

the equation: (5x + 10)° + R = 90°   (2)

R=55° into (2)

(5x + 10)° + 55° = 90°

5x + 10 + 55 = 90

 5x + 65 = 90

5x = 25

x = 5

7 0
3 years ago
The coordinates of △ABC are A(12,8), B(10,18), C(4,16) . After a dilation, the coordinates are A'(6,4), B'(5,9), C'(2,8) A′(6,4)
gogolik [260]
Take note of that every single coordinate had their values divided by 2 after the dilation. Therefore, the scaling factor is 1/2, or .5
7 0
3 years ago
Read 2 more answers
The volume of a right cylinder is V = πr2h. If we have an oblique cylinder, like in the figure, what is the volume of a cross-se
olchik [2.2K]
Since you did not attach any picture we cannot say for sure what is the correct answer, but we can discuss the options in order to find the most probable correct answer.

First of all, according to the Cavalieri's principle, an oblique cylinder has the same volume as a right cylinder with the same base surface area and same height.
A cross-section of an oblique cylinder will be a small right cylinder with the same base surface area and a height as small as possible.

I guess the oblique cylinder has height h and it is divided into many (probably 10) cross-sections.

Option A: <span>πr2h
This is exactly the volume of the right cylinder, therefore, unless you are given a cross-section of height h (which would be too easy), this won't be the correct answer.

Option B: </span><span>4πr2h
This is 4 times the right cylinder. Again, here the height of the cross-section should</span> be 4h, but it doesn't sound like a possible data (too easy again).

Option C: <span>1 10 πr2h
Here comes a n issue with the notation: I think the right number you meant to write is (1/10)</span>·πr2h and not 110·<span>πr2h.
If I am right, this means that your oblique cylinder of height h is divided into 10 cross-sections, and therefore the volume of each of these cross-sections will be a tenth of the volume of the oblique cylinder, which means </span>1/10·<span>πr2h.

Option D: </span><span>1 2 πr2h
Here, we have the same notation issue as before. I think you meant (1/2)</span>·<span>πr2h.
Here, your oblique cylinder height h should be divided into only 2 cross-sections. Now, we said the cross-section's height should be the smallest as possible, so an oblique cylinder divided only into two pieces doesn't sound good.

Therefore, the most probable correct answer will be C) </span>(1/10)·<span>πr2h</span>
8 0
3 years ago
Read 2 more answers
Use these functions to answer the questions.
Elenna [48]
Part A
We have f\left(x\right)=\left(x+3\right)^2-1. To solve for the x-intercept, we set f(x) equal to 0.  That is
     \left(x+3\right)^2-1=0

     \left(x+3\right)^2=1

Take the square root of both sides, 
     x+3=1

     x=-2

The x-intercept is (-2,0). 

To solve for the y-intercept, we set x=0. That is 
     y=\left(0+3\right)^2-1=3^2-1=9-1=8

The y-intercept is (0, 8) 

The coordinates of the optimum point are actually the vertex which can be easily seen from the vertex form equation given above. The minimum point is (-3, -1).

Part B.
We have g\left(x\right)=-2x^2+8x+3.
Factor out -2
     =-2\left(x^2-4x\right)+3

Complete the square
     =-2\left(x^2-4x+4\right)+3-2\left(4\right)

Simplify
     g(x)=-2\left(x-2\right)^2-5

Part C
We have h\left(t\right)=-16t^2+28t.
The maximum height is 12.25 feet after 0.875 seconds from the time of the jump. The dolphin will be back in the water after 1.75 seconds. The graph of the jump is shown in the photo. 
     
    

4 0
3 years ago
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