All categories

3 years ago

100 POINTS! SOLVE FOR X. PIC ATTACHED. Random answers will be reported.

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

8 0

Answer:

x = 29

Step-by-step explanation:

The angles (2x- 10) and (4x+ 16) are consecutive interior angles and therefore are corresponding

<u>Corresponding angles sum to 180</u>

(2x - 10) + (4x + 16) = 180
6x + 6 = 180
6x = 174
x = 174/6
x = 29

You might be interested in

rained 2.9 inches on Monday. On Tuesday, it rained 0.56 inches less than on Monday. How much did it rain on Tuesday?

Gnom [1K]

Answer:

13/20 should be the answer im sorry if it isnt

Step-by-step explanation:

6 0

4 years ago

Help express y in terms of x

Oliga [24]

Answer:

the answer is y=x+2

Step-by-step explanation:

The row of numbers beside the x each are represented by x in the equation. The equation must be true for all numbers. You must look at each number and find the pattern.

1+2= 3

2+2=4

3+2=5

each x value fits the pattern of being equal to the y value below it when adding two.

hope this helps

6 0

3 years ago

Don't get it at all I am have lots of trouble

maks197457 [2]

47. B 9
48, D 81
49. A 5
50 C 45

5 0

3 years ago

What is 0.5% as a decimal

g100num [7]

Answer:

0.005

Step-by-step explanation:

To change a percent to a decimal move the decimal point two places to the left.

Please mark the brainliest

8 0

4 years ago

A child wanders slowly down a circular staircase from the top of a tower. With x,y,zx,y,z in feet and the origin at the base of

babymother [125]

Answer:

a) The tower is 90 feet tall

b) She reaches the bottom at t = 18 minutes.

c) Her speed at time t is 5 \sqrt[]{5} ft/minute

d) Her acceleration at time t is 10 ft/minute^2

Step-by-step explanation:

Consider the path described by the child as going down the tower to have the following parametrization $\gamma(t) = (10\cos t, 10 \sin t, 90-5t)$

a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.

b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.

c) Given the parametrization we are given, the velocity of the child at time t is given by $\frac{d\gamma}{dt}= (\frac{d}{dt}(10\cos t), \frac{d}{dt} (10 \sin t ), \frac{d}{dt}(90-5t)) = (-10 \sin t, 10 \cos t, -5)$ . The speed is defined as the norm of the velocity vector,

so, the speed at time t is given by $v = \sqrt[]{(-10 \sin t)^2+(10 \cos t)^2+(-5)^2} = \sqrt[]{100(\sin^2 t + \cos^2 t)+25} = \sqrt[]{125}= 5 \sqrt[]{5}$

d) ON the same fashion we want to know the norm of the second derivative of $\gamma$ .

We have that $\gamma ^{''}(t) =(-10\cost t, -10 \sin t , 0)$ so the acceleration is given by $\sqrt[]{100(\cos^2 t+ \sin^2 t )} = 10$

6 0

3 years ago