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just olya [345]
3 years ago
9

Find the equation of a line that bisects the line segment joining the points (-1, 3) and (5, -3) and intersects the x-axis at th

e origin.
Mathematics
1 answer:
maria [59]3 years ago
7 0

Answer:

x - axis or y = 0

Step-by-step explanation:

Bisector passes through the midpoint

(-1+5)/2, (3-3)/2

(2,0)

Line passing through (2,0) and the origin is the x-axis or y = 0

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-9x-6y=15,9x-10y=145
vazorg [7]

-9x-6y=15

9x-10y=145

Find y first by eliminating the x.

Add the two equations together.

-9x+9x-6y+-10y=15+145

-6y+-10y=15+145

-6y-10y=160

-16y=160

divide both sides by -16 to get +y

-16y/-16=160/-16

y=-10

Use the substitution method to find x

-9x-6y=15

-9x-6(-10)=15

Do the bracket first

-9x+60=15

Move +60 to the other side. Sign changes from +60 to -60.

-9x+60-60=15-60

-9x=-45

Divide both sides by -9

-9x/-9=-45/-9

x=5

Answer:

( 5,-10)

3 0
3 years ago
3 red balls, 4 white balls, and 1 green ball are in a bag. Balls are drawn without replacement. Find the probability that the re
jolli1 [7]

The probability that the red balls can be drawn before the green ball is drawn is 3/8.

<h3>What is Probability?</h3>

The probability is the measure of the likelihood of an event to happen.

Given:

red balls= 3

white balls = 4

green balls = 1

So, Probability of red balls

=3/8

Probability of green balls

=1/7

Learn more about probability here:

brainly.com/question/14546377

#SPJ1

7 0
2 years ago
Which of the following would eliminate the variable on the left side of the given equation? 36 - 19w = -6w + 41.  1. add -19w  2
Paladinen [302]
W=5/13 you just eliminate the negative sign
5 0
3 years ago
If 2tanA=3tanB then prove that,<br>tan(A+B)= 5sin2B/5cos2B-1​
Fed [463]

By definition of tangent,

tan(A + B) = sin(A + B) / cos(A + B)

Using the angle sum identities for sine and cosine,

sin(x + y) = sin(x) cos(y) + cos(x) sin(y)

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

yields

tan(A + B) = (sin(A) cos(B) + cos(A) sin(B)) / (cos(A) cos(B) - sin(A) sin(B))

Multiplying the right side by 1/(cos(A) cos(B)) uniformly gives

tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A) tan(B))

Since 2 tan(A) = 3 tan(B), it follows that

tan(A + B) = (3/2 tan(B) + tan(B)) / (1 - 3/2 tan²(B))

… = 5 tan(B) / (2 - 3 tan²(B))

Putting everything back in terms of sin and cos gives

tan(A + B) = (5 sin(B)/cos(B)) / (2 - 3 sin²(B)/cos²(B))

Multiplying uniformly by cos²(B) gives

tan(A + B) = 5 sin(B) cos(B) / (2 cos²(B) - 3 sin²(B))

Recall the double angle identities for sin and cos:

sin(2x) = 2 sin(x) cos(x)

cos(2x) = cos²(x) - sin²(x)

and multiplying uniformly by 2, we find that

tan(A + B) = 10 sin(B) cos(B) / (4 cos²(B) - 6 sin²(B))

… = 10 sin(B) cos(B) / (4 (cos²(B) - sin²(B)) - 2 sin²(B))

… = 5 sin(2B) / (4 cos(2B) - 2 sin²(B))

The Pythagorean identity,

cos²(x) + sin²(x) = 1

lets us rewrite the double angle identity for cos as

cos(2x) = 1 - 2 sin²(x)

so it follows that

tan(A + B) = 5 sin(2B) / (4 cos(2B) + 1 - 2 sin²(B) - 1)

… = 5 sin(2B) / (4 cos(2B) + cos(2B) - 1)

… = 5 sin(2B) / (4 cos(2B) - 1)

as required.

5 0
2 years ago
In ΔАВС, m∠ACB = 90°, <br> CD<br> ⊥ <br> AB<br> m∠ACD = 30°, and AC = 6 cm. Find BD
myrzilka [38]
It is 9 because you have a 30,60,90 triangle, and then you can use pythagorean theorem.
3 0
3 years ago
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