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cestrela7 [59]
2 years ago
9

In the following diagram BC is parallel to DE. What is the mesure of x?

Mathematics
2 answers:
Advocard [28]2 years ago
6 0

Answer:

OK wait I have no clue I would be able to help you but you have to put in the "Diagram" so I can solve for all angles, anyways it's been 10 minutes but I found where this question came from: Khan Academy 8th grade illustrative unit 1 lesson 16 Parallel lines and the angles in a triangle, the graph I see has an answer of 37

Step-by-step explanation:

Answer is correct on Khan Academy:

Solve for angle c 41+102=143 180 degrees in a full triangle so you need to do

180-143 to get 37 NOW 37 is the right answer I have put in a lot of effort to get this answer to you anyways to prove it is right 3 ways I will test it by finding the value of every angle and here is the math for that 37+37+102+41+143

A=143 because 102+41 angle DB is 43 because there is one line intersecting the two parallel lines, this means that the angle is also using the same two lines is also 41.

Then 37 was the only two numbers that could be added to get the full 360.

Answer: 37

Here is a screenshot =D  

This was a lot of effort for one 8th grader so if this is sloppy please forgive me.

soldi70 [24.7K]2 years ago
4 0

Answer:

Step-by-step explanation:

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is it possible to draw a tringle with 90 Angle and one leg that is 8 centimeters long and one leg that is 6 centimeters long? is
Delvig [45]
Yes it is The hypotenuse  is  10 cms long.
Because its a right angled triangle ( 90 angle)  the hypotenuse will be 10 cms long - it will obey the Pythagoras theorem:-

10^2 = 8^2 + 6^2

There is only one  right angled triangle with these measurements but there are many  of them which have sides in same ratio

for example 

3, 4 and 5

9,12 and 15

12, 16 and 20  and so on..
5 0
3 years ago
Rodger has 47 cars . Can he group the cars in more than two ways?
deff fn [24]
First of all, Rodger must be making some serious cash to have 47 cars. Going back to the question, he can not group the cars in more than 2 ways. He can only group it in 1 way, 47 groups of 1. Pls brainliest
4 0
2 years ago
4. What is the solution of the system of equations?<br> y=-2x+5<br> y=-2x+20
Norma-Jean [14]

The system of equation, y = -2x + 5 and y = -2x + 20, have no solution.

<h3>How to Find the Solution of a System of Equations?</h3>

The solution to a system of equations is the ordered pair of x and y that makes both equations true.

Given the system of equations,

y = -2x + 5 --> equation 1

y = -2x + 20 --> equation 2

Subtract to eliminate variable "y"

0 = 0 + 25

0 = 25 [this is not true]

Therefore, no solution exist for the system of equations.

Thus, we can conclude that, the system of equation, y = -2x + 5 and y = -2x + 20, have no solution.

Learn more about system of equations on:

brainly.com/question/14323743

#SPJ1

3 0
1 year ago
Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0
Leokris [45]

Substitute v(x)=-2x+y(x), so that \dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}. Then the ODE is equivalent to

\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7

which is separable as

\dfrac{\mathrm dv}{v^2-9}=\mathrm dx

Split the left side into partial fractions,

\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)

so that integrating both sides is trivial and we get

\dfrac{\ln|v-3|-\ln|v+3|}6=x+C

\ln\left|\dfrac{v-3}{v+3}\right|=6x+C

\dfrac{v-3}{v+3}=Ce^{6x}

\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}

\dfrac6{v+3}=1-Ce^{6x}

v=\dfrac6{1-Ce^{6x}}-3

-2x+y=\dfrac6{1-Ce^{6x}}-3

y=2x+\dfrac6{1-Ce^{6x}}-3

Given the initial condition y(0)=0, we find

0=\dfrac6{1-C}-3\implies C=-1

so that the ODE has the particular solution,

\boxed{y=2x+\dfrac6{1+e^{6x}}-3}

5 0
3 years ago
PLZ HELP ASAP
AleksAgata [21]
2075-1195=880$ is the amount of increase

880/1195=0,74 now you only need to multiply with 100
0,74*100=74%
So  the price increased by 74%
7 0
3 years ago
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