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Blizzard [7]
3 years ago
14

Help me on this plzzzz

Mathematics
2 answers:
AleksAgata [21]3 years ago
8 0
2x - 6 = x + 7
x = 13

2x - 6 = 2(13) - 6 = 20

QR = 2(20) = 40

answer C. 40
Blizzard [7]3 years ago
6 0
The Correct Answer Is A.

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What is the measure of Zc<br>​
maks197457 [2]

Answer: 56 degrees

Step-by-step explanation:

We know that angle between b and c is 90 degrees. Because there is a line dividing angles a and b from 124 and c, we also know that each side is 180 degrees (a and b add to 180, and 124 and c add to 180).

124 and c are supplementary angles. We can represent this in an equation to solve for c:

124 + c = 180\\\\c = 56

3 0
2 years ago
What is 0.84 10 times as much as
Aleks [24]
0.84 is 10 times as much as 0.084
5 0
3 years ago
Read 2 more answers
Jake says adding 0 to an addendum does not change a sum. Is he correct
siniylev [52]
Nothing changes if you don't add anything. 

Example:

10+10=20
20+0=20
 
nothing changes.
5 0
2 years ago
Find the point(s) of intersection (if any) of the plane and the line. Also, determine whether the line lies in the plane. 2x - 2
s2008m [1.1K]

Answer:

the intersection is the point P=(x,y,z)=(8,9,14) and the line does not lie in the plane

Step-by-step explanation:

from the equation of the line

x - 1/2 = y + (3/2)/-1 = (z + 1) / 2  = t (parameter)

then the parametric equation of the line is

x= 1/2 +t

y =  (3/2) + t

z = (-1) + 2*t

therefore from the equation of the plane

2x - 2y + z = 12

2*(1/2 +t) - 2[(3/2) + t]  + [(-1) + 2*t]  = 12

1+2*t - 3 -2*t -1 + 2*t = 12

-3 + 2*t = 12

t= 15/2

therefore there is only one intersection of the line with the plane ( then the line does not lie in the plane , since there would be infinite intersection points). The intersection is

x= 1/2 +t = 1/2 +15/2 = 8

y =  (3/2) + t = (3/2) + 15/2 = 9

z = (-1) + 2*t =  (-1) + 2*15/2  = 14

thus the intersection point P=(x,y,z)=(8,9,14)

7 0
2 years ago
Number 2 I need help on how to solve it
Hunter-Best [27]

Let's simplify the following expression:

\text{ }\frac{\text{ \lparen2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{\text{ 6x}^5\text{y}^{-4}\text{z}^{-5}}{\text{ \lparen2x}^6\text{y}^3\text{z}^{-3})^2}

We get,

\text{ }\frac{(\text{2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{  }\frac{6x^5y^{-4}z^{-5}}{(2x^6y^3z^{-3})^2}\text{ }\frac{(2x^4-y^6z^{-4})^2(6x^5y^{-4}z^{-5})}{(2x^2z^3)(2x^6y^3z^{-3})^2}\text{ }\frac{(4x^8\text{ - 4x}^4y^6z^{-4}\text{ + y}^{12}z^{-8})(6x^5y^{-4}z^{-5})}{(2x^2z^3)(4x^{12}y^6z^{-6})}\text{ }\frac{24x^{13}y^{-4}z^{-5}\text{ - 24x}^9y^2z^{-9}\text{ + 6x}^5y^8z^{-13}}{8x^{14}y^6z^{-3}}

4 0
1 year ago
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