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Lena [83]
3 years ago
12

The coordinates of triangle ABC are A(−4, 3), B(2, 3), and C(4, −1). A line segment runs through the triangle with endpoints Y(0

, 1) and Z(3, 1). If triangle ABC is dilated by a scale factor of 2 creating the new triangle of A′B′C′, what can you say about the length of line segment Y′Z′?
Mathematics
1 answer:
-BARSIC- [3]3 years ago
4 0

We can say that the length of line segment Y'Z' will increase by a scale factor of 2.

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Solve for 5 points, will give brainlist
Elanso [62]

Answer:

i know that the first one is not equivlent but i dont know about the others

Step-by-step explanation:

7 0
3 years ago
QR = 36 ; PR = 45 ; PQ = 21, Sin __?___ = 21 / 45
sammy [17]

Answer:

Sin = PQ / PR = OPPOSITE / HYPOTENUSE

Step-by-step explanation:

Please find the attached file for a clearer understanding.

SOHCAHTOA

Sin = opposite / Hypotenuse

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7 0
3 years ago
Read 2 more answers
If the current temperature is -11°F, what change in temperature would result in a final temperature of -27°F?
kirill [66]
-7 F I think is the answers
4 0
3 years ago
can anyone help me with these? I'm having a lot of trouble and I'm extremely behind... I need to catch up badly!!! i have been a
TiliK225 [7]

Answer:

Step-by-step explanation:

5 and 6 are the same, so I will do one of those and you can use the example to do the other one.  

Slope-intercept form is y = mx + b.  If we have 2 points and nothing else, we can use those 2 points in the slope formula to find the slope (the m value in the equation) and then plug that in with either one of the points to get the equation.  The slope formula is:

m=\frac{y_{2} -y_{1} }{x_{2}-x_{1}  }

Using our points in question 5:  (2, 4) (5, 4)

m=\frac{4-4}{5-2} =\frac{0}{3}=0 So m = 0.  Now pick a point and use it in the equation along with the m value to solve for b.  I will use (2, 4):

4 = 0(2)+b and

4 = 0 + b so

b = 4.  

Now we have m = 0 and b = 4 so we fill in the equation with that info:

y = 0x + 4 or simplifying,

y = 4

For question 7 they want the line parallel to y = 3x + 6 that goes through (-10, 2.5).  For a line to be parallel to another line, their slopes have to be identical.  The slope in y = 3x + 6 is 3.  So the slope of the "new" line is going to be 3 as well.  Now we will use that slope and the given point to solve for b, just like in #5:

2.5 = 3(-10) + b and

2.5 = -30 + b so

b = 32.5

Now we will fill in:

y = 3x + 32.5

For question 8 they want the line perpendicular to y = -4x - 2 that goes through (-16, -11).  For a line to be perpendicular to another line, their slopes have to be opposite reciprocals.  Opposite meaning the sign is opposite (positive becomes negative and negative becomes positive), and reciprocal meaning the fraction is flipped upside down.  The slope in our line is -4.  That means that the perpendicular slope is positive 1/4.  Using that along with our point, we will again solve for b:

-11=\frac{1}{4}(-16)+b which simplifies to

-11 = -4 + b so

b = -7

Now we fill in:

y=\frac{1}{4}x-7

For question 8, in order to find the line parallel to the given line, you need to know the slope.  In the form it is currently in, we do not know the slope.  We need to put it into slope-intercept form to find the slope, then we will proceeed as above.  If

x + 4y = 6, then

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(The 3/2 is reduced from 6/4)

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5=(-\frac{1}{4})(-8)+b which simplifies to

5 = 2 + b so

b = 3

Now we fill in, keeping in mind that lines are parallel when they have the exact same slope:

y=-\frac{1}{4}x+3

6 0
3 years ago
Help with the following problem please
Effectus [21]
Given:
The largest circle has a radius of R=7 units.
Let x be the radius of the large shaded circle.
The small shaded circles have a radius of 1/5 of the large shaded circle.
=> the small shaded circles have a radius of r=x/5

By adding up radii, we have the equation
2(r+x+r)=2(x/5+x+x/5)=2R=2*7=14
Simplify:
7x/5=14/2
x=5
=> r=1

Area of outer circle                  = \pi R^2=\pi7^2=49\pi
Area of large shaded circle      = \pi x^2=\pi5^2=25\pi
Area of 4 small shaded circles = 4\pi r^2=4\pi1^2=4\pi
Total area of shaded circles     =25\pi+4\pi=29\pi

Shaded area as a fraction of that of the outer circle 
=\frac{29\pi}{49\pi}=29/49

3 0
3 years ago
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