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2 years ago

Geometry help please

Mathematics

1 answer:

Law Incorporation [45]2 years ago

3 0

<h3>Answer: RJ = 10</h3>

=======================================================

Explanation:

Recall that midsegments are half as long as the side they are parallel to. In this case, midsegment PQ is parallel to segment GJ. This means PQ is half as long as GJ. Phrased another way, we can say GJ is twice as long as PQ.

So,

GJ = 2*PQ

Also, we can see that GR = RJ due to the triple tickmarks they share. This leads to GJ = GR+RJ = RJ+RJ = 2*RJ

Or in short,

GJ = 2*RJ

---------

In summary we can see that

GJ = 2RJ
GJ = 2PQ

equating both right hand sides leads to

2RJ = 2PQ

RJ = PQ

Because PQ is 10 units, this means RJ is also 10 units.

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Another way to see why PQ = RJ is to notice how triangles HPQ and QRJ are congruent triangles (use SAS to prove this). The corresponding pieces PQ and RJ are the same length, so PQ = RJ = 10.

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X - 10 = 25 please solve this hurry

Kitty [74]

X - 10 = 25
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2 years ago

Read 2 more answers

Any 10th grader solve it <br>for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$\neq 0$

ie., $RHS\neq 0$

Therefore $LHS\neq RHS$

ie., $\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$

Hence proved

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2 years ago

Solve for x:<br> 3/ X - 4 = 6/ X+1

Lina20 [59]

Answer:

-3/5

Step-by-step exp:

given in the file. Hope it helps.

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2 years ago

A tree farmer Planet 3 types of trees on 22 acres of land .he planted 48 trees per acre.what was the total number of trees the f

Alja [10]

Answer:

The total number of trees the farmer planted = 1056 trees

Step-by-step explanation:

It is given that,

A tree farmer Planet 3 types of trees on 22 acres of land and,

he planted 48 trees per acre

<u>To find total number of plants</u>

There are 22 acres of land.

In one acre he planted 48 trees.

Therefore total number of plants in 22 acers = 22 * 48 = 1056 trees

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2 years ago

Please help this is timed! I’ll give brainliest.

lakkis [162]

Answer:

a. Function 1

b. Function 3

c. Function 2, Function 3 and Function 4

Step-by-step explanation:

✔️Function 1:

y-intercept = -3 (the point where the line cuts across the y-axis)

Slope, using the two points (0, -3) and (1, 2):

$\frac{y_2 - y_1}{x_2 -x_1} = \frac{2 -(-3)}{1 - 0} = \frac{5}{1} = 5$

Slope = 5

✔️Function 2:

y-intercept = -1 (the value of y when x = 0)

Slope, using the two points (0, -1) and (1, -4):

$\frac{y_2 - y_1}{x_2 -x_1} = \frac{-4 -(-1)}{1 - 0} = \frac{-3}{1} = -3$

Slope = -3

✔️Function 3: y = 2x + 5

y-intercept (b) = 5

Slope (m) = 2

✔️Function 4:

y-intercept = 2

Slope = -1

Thus, the following conclusions can be made:

a. The function's graph that is steepest is the function whose absolute value of its slope is greater. Therefore Function 1 is the steepest with slope of 5

b. Function 3 has a y-intercept of 5, which is the farthest from 0.

c. Function 2, Function 3, and Function 4 all have y-intercept that is greater than -2.

-1, 5, and 2 are all greater than -2.

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3 years ago