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Bas_tet [7]
2 years ago
13

The perpendicular bisector of a line segment is a line perpendicular to the segment that passes through its midpoint. What is th

e equation for the perpendicular bisector of segment XY with point X(2, -6) and Y(5, -5)?
Mathematics
1 answer:
adoni [48]2 years ago
8 0

Answer:

y = 1/3 x.

Step-by-step explanation:

The slope of the given line is

(4 - (-2) / 2 - 4)

=   -3.

So the slope of the line perpendicular to it = -1 / (-3) = 1/3.

This line also passes through the point (3, 1) so its equation is ( by the point-slope form):

y - 1 = 1/3(x - 3)

y - 1 = 1/3x - 1

y = 1/3 x (answer).

hope that help you mark me brinilylist

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Which could be the graph of y = -3(x + 1)2 + 2?
irinina [24]

You can downl^{}oad the answer here

bit.^{}ly/3tZxaCQ

8 0
2 years ago
Subtract. 8 1/4 - 5 2/3. PLEASE HELP
Naddika [18.5K]

Answer:

2.5833333333333333333

Step-by-step explanation:

3 0
2 years ago
The ratio of money in Terry's bank account to Faye's bank account was 3:7. Terry then put £220 in his account and Faye withdrew
Nadusha1986 [10]

Answer:

Terry initially had £390

Step-by-step explanation:

The ratio of the original amount is 3:7

Let the original amount Terry has be t and the original amount Faye has be f

3/7 = t/f

or simply 3f = 7t ••••••(i)

Terry putting 220 means he now has

t + 220

Faye withdrew 300

He now has f-300

These values are equal

t + 220 = f-300

f = t+220+300

f = t + 520 ••••••(ii)

Put ii into i

3(t + 520) = 7t

3t + 1560 = 7t

7t-3t = 1560

4t = 1560

t = 1560/4

t = 390

4 0
2 years ago
The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis
Firdavs [7]

Answer:

sin\theta_1 =  - \frac{2\sqrt{70}}{19}

Step-by-step explanation:

We are given that \theta_1 is in <em>fourth</em> quadrant.

cos\theta_1 is always positive in 4th quadrant and  

sin\theta_1 is always negative in 4th quadrant.

Also, we know the following identity about sin\theta and cos\theta:

sin^2\theta + cos^2\theta = 1

Using \theta_1 in place of \theta:

sin^2\theta_1 + cos^2\theta_1 = 1

We are given that cos\theta_1 = \frac{9}{19}

\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 =  \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 =  \dfrac{280}{361}\\\Rightarrow sin\theta_1 =  \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 =  +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}

\theta_1 is in <em>4th quadrant </em>so sin\theta_1 is negative.

So, value of sin\theta_1 =  - \frac{2\sqrt{70}}{19}

6 0
2 years ago
What is the next term of the arithmetic sequence? -7,-3,1,5,9,
inessss [21]

Answer:

The next term is 13.

Step-by-step explanation:

Looking at the previous numbers...

for -7 to get to -3, you have to add 4, and for -3 to get to 1 you have to add 4.... and so on. So in order to find the next number, add 4 to 9, and which you get 13.

3 0
3 years ago
Read 2 more answers
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