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2 years ago

11

Triangle ABC is congruent to triangle PQR, angle B equals 3v+4, and angle Q equals 8v-6, find the measurements of angles B, and

Q. please show step by step, thank you.

1 answer:

Nady [450]2 years ago

7 0

The angels of equality in b question

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Which expression is equal to<br> -18b^-4

victus00 [196]

Answer:

$- \frac{18}{ {b}^{4} }$

Step-by-step explanation:

$- 18b {}^{ - 4}$

express b^-4 with a positive exponent using:

$a {}^{ - n} = \frac{1}{ {a}^{n} }$

$- 18 \frac{1}{ {b}^{4} }$

$- \frac{18}{ {b}^{4} }$

8 0

2 years ago

Please help and explain why, thank you

GalinKa [24]

Answer:

Good morning all friend

4 0

2 years ago

GalinKa [24]

Answer:

y=2x-4

I think this is right

Step-by-step explanation:

7 0

3 years ago

IOr — 5г + 3r — 8г i need help

Anni [7]

Answer:

0r

Step-by-step explanation:

Since they all have r you can pretend it s not there and add it on at the end.

Now 10 - 5 is 5. 5 + 3 is 8. 8 - 8 is 0. tack the r onto the end and you get 0r.

3 0

2 years ago

Using the given information, give the vertex form equation of each parabola.

Amanda [17]

Answer:

The equation of parabola is given by : $(x-4) = \frac{-1}{3}(y+3)^{2}$

Step-by-step explanation:

Given that vertex and focus of parabola are

Vertex: (4,-3)

Focus:( $\frac{47}{12}$ ,-3)

The general equation of parabola is given by.

$(x-h)^{2} = 4p(y-k)$ , When x-componet of focus and Vertex is same

$(x-h) = 4p(y-k)^{2}$ , When y-componet of focus and Vertex is same

where Vertex: (h,k)

and p is distance between vertex and focus

The distance between two points is given by :

L= $\sqrt{(X2-X1)^{2}+(Y2-Y1)^{2}}$

For value of p:

p= $\sqrt{(X2-X1)^{2}+(Y2-Y1)^{2}}$

p= $\sqrt{(4-\frac{47}{12})^{2}+((-3)-(-3))^{2}}$

p= $\sqrt{(\frac{1}{12})^{2}}$

p= $\frac{1}{12}$ and p= $\frac{-1}{12}$

Since, Focus is left side of the vertex,

p= $\frac{-1}{12}$ is required value

Replacing value in general equation of parabola,

Vertex: (h,k)=(4,-3)

p= $\frac{-1}{12}$

$(x-h) = 4p(y-k)^{2}$

$(x-4) = 4(\frac{-1}{12})(y+3)^{2}$

$(x-4) = \frac{-1}{3}(y+3)^{2}$

8 0

2 years ago