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4 years ago

Someone help me on this one pls!!

Mathematics

1 answer:

Nataly [62]4 years ago

7 0

The formula of a midpoint of AB:

$M_{AB}\left(\dfrac{x_A+x_B}{2};\ \dfrac{y_A+y_B}{2}\right)$

We have:

G(b; 0); F(3b; 2b)

substitute:

$M_{GF}\left(\dfrac{b+3b}{2};\ \dfrac{0+2b}{2}\right)\to M_{GF}(2b;\ b)$

Answer: (2b, b)

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Can someone help me with these 4 geometry questions? Pls it’s urgent, So ASAP!!!!

blagie [28]

Question 4

1) $\overline{BD}$ bisects $\angle ABC$ , $\overline{EF} \perp \overline{AB}$ , and $\overline{EG} \perp \overline{BC}$ (given)

2) $\angle FBE \cong \angle GBE$ (an angle bisector splits an angle into two congruent parts)

3) $\angle BFE$ and $\angle BGE$ are right angles (perpendicular lines form right angles)

4) $\triangle BFE$ and $\triangle BGE$ are right triangles (a triangle with a right angle is a right triangle)

5) $\overline{BE} \cong \overline{BE}$ (reflexive property)

6) $\triangle BFE \cong \triangle BGE$ (HA)

Question 5

1) $\angle AXO$ and $\angle BYO$ are right angles, $\angle A \cong \angle B$ , $O$ is the midpoint of $\overline{AB}$ (given)

2) $\triangle AXO$ and $\triangle BYO$ are right triangles (a triangle with a right angle is a right triangle)

3) $\overline{AO} \cong \overline{OB}$ (a midpoint splits a segment into two congruent parts)

4) $\triangle AXO \cong \triangle BYO$ (HA)

5) $\overline{OX} \cong \overline{OY}$ (CPCTC)

Question 6

1) $\angle B$ and $\angle D$ are right angles, $\overline{AC}$ bisects $\angle BAD$ (given)

2) $\overline{AC} \cong \overline{AC}$ (reflexive property)

3) $\angle BAC \cong \angle CAD$ (an angle bisector splits an angle into two congruent parts)

4) $\triangle BAC$ and $\triangle CAD$ are right triangles (a triangle with a right angle is a right triangle)

5) $\triangle BAC \cong \triangle DCA$ (HA)

6) $\angle BCA \cong \angle DCA$ (CPCTC)

7) $\overline{CA}$ bisects $\angle ACD$ (if a segment splits an angle into two congruent parts, it is an angle bisector)

Question 7

1) $\angle B$ and $\angle C$ are right angles, $\angle 4 \cong \angle 1$ (given)

2) $\triangle BAD$ and $\triangle CAD$ are right triangles (definition of a right triangle)

3) $\angle 1 \cong \angle 3$ (vertical angles are congruent)

4) $\angle 4 \cong \angle 3$ (transitive property of congruence)

5) $\overline{AD} \cong \overline{AD}$ (reflexive property)

6) $\therefore \triangle BAD \cong \triangle CAD$ (HA theorem)

7) $\angle BDA \cong \angle CDA$ (CPCTC)

8) $\therefore \vec{DA}$ bisects $\angle BDC$ (definition of bisector of an angle)

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2 years ago

What is the distance between the points (-3,4) and (5,4)?

marta [7]

Answer:

8/1

Step-by-step explanation:

You have to go up 8 more to get to (5,4) I hope Im right and this helps

5 0

3 years ago

Read 2 more answers

An envelope contains three cards: a black card that is black on both sides, a white card that is white on both sides, and a mixe

Over [174]

Answer:

There is a 2/3 probability that the other side is also black.

Step-by-step explanation:

Here let B1: Event of picking a card that has a black side

B2: Event of picking a card that has BOTH black side.

Now, by the CONDITIONAL PROBABILITY:

$P(B_2/B_1 ) = \frac{P(B_1\cap B_2)}{P(B_1)}$

Now, as EXACTLY ONE CARD has both sides BLACK in three cards.

⇒ P (B1 ∩ B2) = 1 /3

Also, Out if total 6 sides of cards, 3 are BLACK from one side.

⇒ P (B1 ) = 3 /6 = 1/2

Putting these values in the formula, we get:

$P(B_2/B_1 ) = \frac{P(B_1\cap B_2)}{P(B_1)} = \frac{1}{3} \times\frac{2}{1} = \frac{2}{3}$

⇒ P (B2 / B1) = 2/3

Hence, there is a 2/3 probability that the other side is also black.

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3 years ago

B(x)=|x+4| what is b(-10)

e-lub [12.9K]

Answer:

Step-by-step explanation:

b(-10) = |-10+4| will be |-6|

The absolute value of |-6| is 6.

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3 years ago

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Evaluate the line integral, where C is the given curve. C sin(x) dx + cos(y) dy, where C consists of the top half of the circle

kenny6666 [7]

Parameterize the circular part of $C$ (call it $C_1$ ) by

$x=4\cos t$

$y=4\sin t$

wih $0\le t\le\pi$ , and the linear part (call it $C_2$ ) by

$x=-4-t$

$y=4t$

with $0\le t\le1$ .

Then

$\displaystyle\int_C\sin x\,\mathrm dx+\cos y\,\mathrm dy=\left\{\int_{C_1}+\int_{C_2}\right\}\sin x\,\mathrm dx+\cos y\,\mathrm dy$

$=\displaystyle\int_0^\pi(-4\sin t\sin(4\cos t)+4\cos t\cos(4\sin t))\,\mathrm dt+\int_0^1(-\sin(-4-t)+\cos4t)\,\mathrm dt$

$=0+\displaystyle\int_0^1(\sin(t+4)+\cos4t)\,\mathrm dt$

$=\cos4-\cos5+\dfrac{\sin4}4$

7 0

3 years ago