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1 year ago

Find the value of x in the given figure of a triangle

Mathematics

1 answer:

Luba_88 [7]1 year ago

5 0

In the triangle ABE

step 1

Find out the measure of angle AEB

m by form a linear pair

step 2

Find out the measure of angle ABE

m by alternate interior angles

step 3

Find out the measure of angle x

Remember that

The sum of the interior angles in any triangle must be equal to 180 degrees

msubstitute given values

x+100+30=180

x=180-130

<h2>x=50 degrees</h2>

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4x + 4y = 20<br> 2x - 2y = 2 .

Orlov [11]

Answer: Can you tell me how do you want it answered?

Step-by-step explanation:

5 0

3 years ago

Where is Point B on the number line

pishuonlain [190]

I think it’s -3 1/3 because if you count your lines after -3 to -2 you get 3 and it’s on the first line so 1/3! Hope this makes sense :)

3 0

3 years ago

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-10 + x + 4-5 > 7x - 5

coldgirl [10]

Answer:

x < -1

Step-by-step explanation:

Step 1: Write out inequality

-10 + x + 4 - 5 > 7x - 5

Step 2: Solve

x - 6 - 5 > 7x - 5

x - 11 > 7x - 5

-11 > 6x - 5

-6 > 6x

-1 > x

Step 3: Rewrite

x < -1

3 0

3 years ago

Read 2 more answers

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago

For every 9 black and white copies printed there are 2 color copies printed. If there are 200 color copies printed how many blac

alexandr402 [8]

Answer:

<u>900</u> black and white copies are printed.

Step-by-step explanation:

Given:

For every 9 black and white copies printed there are 2 color copies printed.

If there are 200 color copies printed.

Now, to find the number of black and white copies are printed.

Let the number of black and white copies printed be $x.$

As given, for every 9 black and white copies printed there are 2 color copies printed.

So, 9 is equivalent to 2.

If there are 200 color copies printed.

Thus, $x$ is equivalent to 200.

Now, to solve by using cross multiplication method:

$\frac{9}{2} =\frac{x}{200}$

<em>By cross multiplying we get:</em>

$1800=2x$

<em>Dividing both sides by 2 we get:</em>

$900=x$

$x=900.$

Therefore, 900 black and white copies are printed.

8 0

3 years ago

Read 2 more answers