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2 years ago

Line AB is drawn from A(0,10) to B(-7,-4). Find point C that partitions line AB in ratio 5:2

1 answer:

8 0

C) (-5, 0)

Working;
Use the formula;
x coordinate= $\frac{ k_{1} x_{B}+ k_{2} x_{A} }{ k_{1} + k_{2} }$ $\frac{ k_{1} y_{B} + k_{2} y_{A} }{ k_{1} + k_{2} }$

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Given:m<6=m<8b l l cprove:a l l b

makvit [3.9K]

1) From the data, we can see that

$\angle6=\angle8$

is given.

2) Next,

$\angle6=\angle7$

since they are alternate angles.

3) By substitution, these means that

$\angle7=\angle8$

4) Finally, a || b since

$\angle7=\angle8$

this is because, angle 7 and angle 8 are corresponding angles.

Corresponding angles are angles that are on the same corner at each intersection. For instance, 2 and 6

4 and 8, 1 and 5, 3 and 7

In our case, 7 and 8 are corresponding angles

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5 months ago

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Whats the distance between coordinates (-6,8) and (-6,-3)

insens350 [35]

Answer:

there away from eachother for 5 spaces.

Step-by-step explanation:

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2 years ago

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How to figure out the equation 3/5 / 2/3

posledela

Answer:

2/5

Step-by-step explanation:

3/5 / 2/3

3/5 ÷ 2/3

Convert ÷ sign into multiplication, that will turn 2/3 into 3/2

3/5 X 2/3

Divide by 3

3/5 by 3

1/5

2/3 by 3

2/1

1/5 X 2/1

2/5

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2 years ago

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Lindsey's family is hosting an end-of-summer burger bash. Her parents buy a family-size package of ground beef, and Lindsey uses

hammer [34]

Answer:

5 lbs

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

Well, 20 burgers that each weigh 0.25 lbs is simple to find.

20*0.25=P

{P=pounds}

P=5 lbs

(っ◔◡◔)っ ♥ Hope this helped! Have a great day! :) ♥

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2 years ago

Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to chec

klasskru [66]

Answer:

There is a horizontal tangent at (0,-4)

The tangent is vertical at (-2,-3) and (2,-3).

Step-by-step explanation:

The given function is defined parametrically by the equations:

$x=t^3-3t$

and

$y=t^2-4$

The tangent function is given by:

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\implies \frac{dy}{dx}=\frac{2t}{3t^2-3}$

The tangent is vertical at when $\frac{dx}{dt}=0$

$\implies \frac{3t^2-3}{2t}=0$

$\implies 3t^2-3=0$

$\implies 3t^2=3$

$\implies t^2=1$

$\implies t=\pm1$

When t=1,

$x=1^3-3(1)=-2$ and $y=1^2-4=-3$

When t=-1,

$x=(-1)^3-3(-1)=2$ and $y=(-1)^2-4=-3$

The tangent is vertical at (-2,-3) and (2,-3).

The tangent is horizontal, when $\frac{dy}{dx}=0$ or $\frac{dy}{dt}=0$

$\implies 2t=0$

$\implies t=0$

When t=0,

$x=0^3-3(0)=0$ and $y=0^2-4=-4$

There is a horizontal tangent at (0,-4)

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2 years ago