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2 years ago

Find the measure of angle x in the figure below:

Mathematics

1 answer:

Sergeeva-Olga [200]2 years ago

3 0

Answer:

30 degrees.

Step-by-step explanation:

We are going to use two theorems: Triangle Sum Theorem and Vertical Angles Theorem. Please let me know if you need me to define them.

Here is how I solved this problem:

75+75 = 150.

180-150 = 30(Triangle Sum theorem)

x = 30(Vertical Angles theorem.)

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The midpoint of segment XY is (6, -3). The coordinates of one endpoint are X(-1, 8). Find the coordinates of endpoint Y.

Snezhnost [94]

Answer:

The answer to your question is: Y = (13, -14)

Step-by-step explanation:

Data

midpoint = mp = (6, -3)

one endpoint = X = (-1, 8)

second endpoint = Y = (x, y)

Formula

Xmp = (x1 + x2) / 2

Ymp = (y1 + y2) / 2

Process

x2 = 2xmp - x1

y2 = 2ymp - y1

x2 = 2(6) - (-1) y2 = 2(-3) - 8

x2 = 12 + 1 y2 = -6 - 8

x2 = 13 y2 = - 14

Y = (13, -14)

4 0

3 years ago

How many terms are in the expression 5a+1? 1 4 3 2

Studentka2010 [4]

Answer:

I think the answer is 2

Step-by-step explanation:

7 0

2 years ago

3497 to 1 significant figure

kirza4 [7]

1 significant figure would be 3000

7 0

2 years ago

The height of a ball thrown into the air after t seconds have elapsed is h = −16t2 + 40t + 6 feet. What is the first time, t, wh

sp2606 [1]

<h3>The first time when the ball will reach a height of 20 feet is 0.42 seconds</h3>

Solution:

Given that,

The height of a ball thrown into the air after t seconds have elapsed is:

$h = -16t^2 + 40t + 6$

What is the first time, t, when the ball will reach a height of 20 feet?

Substitute h = 20

$20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0$

Solve by quadractic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=8,\:b=-20,\:c=7$

$t = \frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:8\cdot \:7}}{2\cdot \:8}\\\\t = \frac{20 \pm \sqrt{176}}{16}\\\\t = \frac{20 \pm 4\sqrt{11}}{16}\\\\t = \frac{ 5 \pm \sqrt{11}}{4}\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084$

Rounding off we get,

t = 2.08 , t = 0.42

Thus the first time when the ball will reach a height of 20 feet is 0.42 seconds

4 0

3 years ago

NEED HELP!

diamong [38]

Nope, but if you need help you can tell me what the questions are and I'll try solve them!

6 0

3 years ago

Read 2 more answers