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3 years ago

Find the value of x. Then, find the measure of each labelled angle.

Mathematics

1 answer:

wolverine [178]3 years ago

3 0

Answer:

x = 7deg

Step-by-step explanation:

7x + 10 + 15x + 16 = 180 (angles adjacent to each other in a //polygon are supplementary)

22x = 154

x = 7deg

This is a question on polygons and angles. If you wish to venture further into it/understand this topic better, you may want to follow my Instagram account (learntionary), where I post some of my own notes on certain topics and also some tips that may be useful to you :)

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Victoria’s printer can print 8.804 in one minute if Victoria’s prints page for 0.903 minutes, about how many pages will she have

SVETLANKA909090 [29]

Victoria’s printer can print 8.804 in one minute if Victoria’s prints page for 0.903 minutes,

We apply unitary method

In one minute the number of pages printed is 8.804. we need to find the number of pages for 0.903 minutes

1 minute = 8.804 pages

0.903 minutes = ? pages

We multiply 8.804 by 0.903

8.804 * 0.903 = 7.95

Number of pages printed = 7.95 pages

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3 years ago

Which variable is most important to the following problem? At 9:45 a.m. a patient’s temperature was 101.5 degrees at 10:41, the

statuscvo [17]

Answer:

Option A is the important variable.

Step-by-step explanation:

Given is - At 9:45 a.m. a patient’s temperature was 101.5 degrees at 10:41, the nurse took the patients temperature again and found it was 105.8 degrees, the highest ever recorded.

The variable that is most important here is A: the number of minutes it took for the to reach its peak.

Firstly it is not mentioned if 10:41 is am or pm.

If its am, then such a temperature rise in less than an hour is very dangerous.

So, the important variable is option A.

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3 years ago

The medium in which a story is presented most affects the

katen-ka-za [31]

Answer:

audience's perspective of a story - please give brainliest

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3 years ago

The rainfall for one year was 35.8 inches. What was the approximate rainfall per month

Aloiza [94]

The rainfall for one month would be 2.983 inches per month.

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3 years ago

Read 2 more answers

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

3 years ago