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3 years ago

I don’t understand this and I did take notes on this on the computer but I still don’t understand it.

Mathematics

1 answer:

Serggg [28]3 years ago

3 0

Answer:

I'[−3, −2], Q'[2, 0], ?'[0, 1], E'[−3, −1]

Step-by-step explanation:

The line of reflection is at y = −2, so looking at the line and where the coordinates of the shape are, you will have this:

I[−3, −2] → I'[−3, −2] (on the line, so it remains the same)

Q[2, −4] → Q'[2, 0]

?[0, −5] → ?'[0, 1]

E[−3, −3] → E'[−3, −1]

I am joyous to assist you anytime.

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Find the equation of the line that passes through points (3,5) and (-6,2). Write the equation in slope intercept form.

andrezito [222]

1) gradient of line = Δ y ÷ Δ x
= (5 -2) ÷ (3 - (-6))
= ¹/₃

using the point-slope form (y-y₁) = m(x-x₁)
using (3,5)
(y - 5) = ¹/₃ (x -3)
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⇒ y = ¹/₃ x + 4 [OPTION D]

2) y = 2x + 5 .... (1)
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by substituting y in (1) for y in (2)

2x + 5 = ¹/₂ x + 6
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x = ²/₃

by substituting found x (2)

 y = ¹/₂ (²/₃) + 6
y = ¹⁹/₃

∴ common point is (²/₃ , ¹⁹/₃) thus answer is FALSE [OPTION B]

3) Yes [OPTION A]
This is because the both have a gradient of 5 and if they have the same gradient then that means that the two lines are parallel to each other.

4) No [OPTION B]
Two lines are perpendicular if their gradients multiply to give - 1 and as such one is the negative reciprocal of the other. Since both gradients are ¹/₂ then they are actually parallel and not perpendicular.

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3 years ago

Please can someone help me with this question

sasho [114]

Answer:

r = - $\frac{6a}{1-5m^2}$

Step-by-step explanation:

Rewrite the equation as $\sqrt{x} 6a+r/5r$ = m

Remove the radical on the left side of the equation by squaring both sides of the equation.

( $\sqrt{\frac{6a+r}{5r} )^{2}$ = m^2

Then, you simplify each of the equation.

Rewrite: ( $\sqrt{\frac{6a+r}{5r} )^{2}$ as $\frac{6a+r}{5r} = m^{2}$

Remove any parentheses if needed.

Solve for r.

Multiply each term by r and simplify."

Multiply both sides of the equation by 5.

6a+r= m^2r⋅(5)

Remove parentheses.

Move 5 to the left of (m ^2) r

6a+r=5m^2)r

Subtract 5m^2)r from both sides of the equation.

6a+r-5m^2)r=0

Subtract 6a from both sides of the equation.

r-5m^2)r=-6a

Factor r out of r-5m^2)r

r(1-5m^2)=-6a

Divide each term by 1-5m^2 and simplify.

r = - $\frac{6a}{1-5m^2}$

There you go, hope this helps!

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You can only draw one unique isosceles triangle that contains an angle of 55°?

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