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myrzilka [38]
3 years ago
14

Point Q is on line segment PR. Given QR= 2x + 4, PQ = x, and

Mathematics
1 answer:
34kurt3 years ago
8 0

Answer:

Step-by-step explanation:

Point P is on line segment

O

Q

‾

OQ

​

. Given

O

P

=

6

,

OP=6,

O

Q

=

4

x

−

3

,

OQ=4x−3, and

P

Q

=

3

x

,

PQ=3x, determine the numerical length of

O

Q

‾

.

OQ

​

.

Label known information:

Label known information:

O

P

Q

6

3x

OQ = 4x – 3

O

P

+

P

Q

=

OP+PQ=

 

 

O

Q

OQ

6

+

3

x

=

6+3x=

 

 

4

x

−

3

4x−3

−

4

x

−4x=

 

 

−

4

x

−4x

−

x

+

6

=

−x+6=

 

 

−

3

−3

−

6

−6=

 

 

−

6

−6

−

x

=

−x=

 

 

−

9

−9

−

x

−

1

=

−1

−x

​

=

 

 

−

9

−

1

−1

−9

​

x

=

x=

 

 

9

9

Plug in value of

x

to find

O

Q

:

Plug in value of x to find OQ:

O

Q

=

4

x

−

3

=

4

(

9

)

−

3

=

33

OQ=4x−3=4(9)−3=33

You can plug

x

into each expression:

You can plug x into each expression:

O

P

Q

6

3(9)

OQ = 4(9) – 3

Simplify:

Simplify:

O

P

Q

6

27

OQ = 33

Final Answer:

Final Answer:

O

Q

=

33

OQ=33

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kirill [66]

Answer:

Idaho

Step-by-step explanation:

3 0
3 years ago
1.) Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth .x squared minus 21 x equals n
Andrew [12]
Part 1
We are given x^2-21x=-4x. This can be rewritten as x^2-18x=0.
Therefore, a=1, b=-18, c=0.
Using the quadratic formula
     x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(-18\right)\pm \sqrt{\left(-18\right)^2-4\left(1\right)\left(0\right)}}{2\left(1\right)}
     x=\frac{18\pm 18}{2}

The values of x are
     x_1=\frac{18-18}{2}=0
     x_2=\frac{18+18}{2}=18

Part 2
Since the values of y change drastically for every equal interval of x, the function cannot be linear. Therefore, the kind of function that best suits the given pairs is a quadratic function. 

Part 3.
The first equation is y=x^2+2.
The second equation is y=3x+20.

We have 
     x^2+2=3x+20
     x^2-3x-18=0
Factoring, we have 
     \left(x-6\right)\left(x+3\right)=0
Equating both factors to zero.
     x_1-6=0\rightarrow x_1=6
     x_2+3=0\rightarrow x_2=-3

When the value of x is 6, the value of y is 
     y=3\left(6\right)+20=38

When the value of x is -3, the value of y is 
     y=3\left(-3\right)+20=11

Therefore, the solutions are (6,38) or (-3,11)
7 0
3 years ago
Can some one explain in an easy way pls
Helga [31]
What needs to be done first is to add up females and males that have passed.

42 + 14 = 56

so out of 56 students who passed 42 females passed 42/56 = 3/4 = 0.75

out of 56 students who passed, 14 males passed which turns into 14/56 = 1/4 = 0.25

check work; 0.75 + 0.25 = 1.00

NOW WE ARE DOING FAILS.

15 + 5 = 20

so out of 20 students who failed, 15 females failed so it turns into 15/20 = 3/4 = 0.75

out of 20 students who failed, 5 males failed. 5/20 = 1/4 = 0.25

check work; 0.75 + 0.25 = 1.00

i hope this helped! :)

7 0
3 years ago
Triangle ABC and triangle PQR each have two sides whose lengths are 7 and an angle whose measure is 40. Are the triangles congru
Elena L [17]

No

Step-by-step explanation:

they could be but it doesnt say where the angle is so there isn't enough information to tell if they are congruent. answer is no

8 0
3 years ago
Ivan used coordinate geometry to prove that quadrilateral EFGH is a square.
Gelneren [198K]

Answer:

(A)Segment EF, segment FG, segment GH, and segment EH are congruent

Step-by-step explanation:

<u>Step 1</u>

Quadrilateral EFGH with points E(-2,3), F(1,6), G(4,3), H(1,0)

<u>Step 2</u>

Using the distance formula

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Given E(-2,3), F(1,6)

|EF|=\sqrt{(6-3)^2+(1-(-2))^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}

Given F(1,6), G(4,3)

|FG|=\sqrt{(3-6)^2+(4-1)^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}

Given G(4,3), H(1,0)

|GH|=\sqrt{(0-3)^2+(1-4)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt{2}

Given E (−2, 3), H (1, 0)

|EH|=\sqrt{(0-3)^2+(1-(-2))^2}=\sqrt{(-3)^2+(3)^2}=\sqrt{18}=3\sqrt{2}

<u>Step 3</u>

Segment EF ,E (−2, 3), F (1, 6)

Slope of |EF|=\frac{6-3}{1+2} =\frac{3}{3}=1

Segment GH, G (4, 3), H (1, 0)

Slope of |GH|= \frac{0-3}{1-4} =\frac{-3}{-3}=1

<u>Step 4</u>

Segment EH, E(−2, 3), H (1, 0)

Slope of |EH|= \frac{0-3}{1+2} =\frac{-3}{3}=-1

Segment FG, F (1, 6,) G (4, 3)

Slope of |EH| =\frac{3-6}{4-1} =\frac{-3}{3}=-1

<u>Step 5</u>

Segment EF and segment GH are perpendicular to segment FG.

The slope of segment EF and segment GH is 1. The slope of segment FG is −1.

<u>Step 6</u>

<u>Segment EF, segment FG, segment GH, and segment EH are congruent. </u>

The slope of segment FG and segment EH is −1. The slope of segment GH is 1.

<u>Step 7</u>

All sides are congruent, opposite sides are parallel, and adjacent sides are perpendicular. Quadrilateral EFGH is a square

4 0
3 years ago
Read 2 more answers
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