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3 years ago

Show all work to receive credit What is m

Mathematics

1 answer:

ella [17]3 years ago

8 0

Line MK intersects line JL at point N.
This means that:
measure angle MNJ is equal to measure angle LNK as they are opposite angles.

So,
5x+2 = 3(x+14)
5x+2 = 3x + 42
5x-3x = 42-2
2x = 40
x = 20

Therefore:
measure angle MNJ = 5(20)+2 = 102 degrees
measure angle LNK = 3(20+14) = 102 degrees
You will notice that they are both equal as stated in the beginning.

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Work out the coordinates of the midpoint of MN. M (-5, 3) N(7, -8)

Angelina_Jolie [31]

Answer:

5,3 joules

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2 years ago

Read 2 more answers

If m = 8, what is 4m + 2m? Hint: "4m" means "4 TIMES m" and "2m" means "2 TIMES m"

aleksandr82 [10.1K]

Answer:

Step-by-step explanation:

4m means 4xm, and 2m means 2xm

Now, if m = 8, 4m means 4x8, and 2m means 2x8. (Technically, substituting the number in place of the variable m)

4x8 = 32

2x8 = 16

So the equation now is 32+16

Do the math and you get 48

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3 years ago

Read 2 more answers

A group of 11 people are going to the movies one Saturday. Adult tickets cost $12.50 each and child tickets cost $7.75 each. If

Lostsunrise [7]

The answer is B. 5 !!

3 0

3 years ago

1. In which quadrant or on which axis do each of the points (-2,4) , (3,-1) ,(-1,0), (1,2) and (-3,-5) lie? Verify your answer b

pashok25 [27]

Point (-2 , 4) lies in the 2nd quadrant

Point (3 , -1) lies in the 4th quadrant

Point (-1 , 0) lies on the negative part of the x-axis

Point (1 , 2) lies in the 1st quadrant

Point (-3 , -5) lies in the 3rd quadrant

Step-by-step explanation:

Let us revise the signs of the coordinates in each quadrant

x and y coordinates are positive in the 1st quadrant
x-coordinate is negative and y-coordinate is positive in the 2nd quadrant
x and y coordinates are negative in the 3rd quadrant
x-coordinate is positive and y-coordinate is negative in the 4th quadrant
The general point on the positive part of x-axis is (x , 0), and on the negative part of x-axis is (-x , 0)
The general point on the positive part of y-axis is (0 , y), and on the negative part of y-axis is (0 , -y)

Now let us check the points

Point (-2 , 4) ⇒ red point

∵ x = -2 and y = 4

∵ x-coordinate is negative and y-coordinate is positive

∴ Point (-2 , 4) lies in the 2nd quadrant

Point (3 , -1) ⇒ blue point

∵ x = 3 and y = -1

∵ x-coordinate is positive and y-coordinate is negative

∴ Point (3 , -1) lies in the 4th quadrant

Point (-1 , 0) ⇒ green point

∵ x = -1 and y = 0

∵ x-coordinate is negative and y-coordinate is zero

∴ Point (-1 , 0) lies on the negative part of the x-axis

Point (1 , 2) ⇒ purple point

∵ x = 1 and y =2

∵ x and y coordinates are positive

∴ Point (1 , 2) lies in the 1st quadrant

Point (-3 , -5) ⇒ black point

∵ x = -3 and y = -5

∵ x and y coordinates are negative

∴ Point (-3 , -5) lies in the 3rd quadrant

Look to the attached graph for more understand

Learn more:

You can learn more about the four quadrant in brainly.com/question/9381523

#LearnwithBrainly

5 0

3 years ago

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist

yuradex [85]

Answer:

0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.

Step-by-step explanation:

Given the differential equation

(t - 5)y' + (ln t)y = 6t

and the condition y(1) = 6

We can rewrite the differential equation by dividing it by (t - 5) as

y' + [(ln t)/(t - 5)]y = 6t/(t - 5)

(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).

6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)

We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).

But the initial condition is y = 6, when t = 1.

The solution to differential equation is certain to exist at (0, 5)

Which implies that

0 < t < 5

is the required interval.

3 0

3 years ago