-10=x/2+6<br> Answers <br> -8 -14 -32 18

Angle AOC has what measurement according to the protractor? A. 150° B. 50° C. 30° D. 130°

maxonik [38]

Answer:

∠AOC = 50°

B is correct.

Step-by-step explanation:

In the given figure of protractor measure the ∠AOC

Please find the attachment of protractor.

OA is horizontal line which is base of protractor.

In angle AOC, two legs OA and OC

OA is base of protractor. OC another leg.

Now we see position of leg OC of angle AOC.

∠AOC is acute angle. So, see the number bottom of protractor.

Hence, the measure of ∠AOC is 50°

7 0

3 years ago

Read 2 more answers

(4 x 10^2) + 120.5<br> What’s the answer to this

erastovalidia [21]

Answer:

520.5

Step-by-step explanation:

5 0

2 years ago

Use the given minimum and maximum data entries, and the number of classes, to find the class width, the lower class limits,

Gennadij [26K]

Using proportions and the information given, it is found that:

The class width is of 14.375.

The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.

The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

-------------------------

Minimum value is 19.
Maximum value is of 134.
There are 8 classes.
The classes are all of equal width, thus the width is of:

$W = \frac{134 - 19}{8} = 14.375$

-------------------------

The intervals will be of:

19 - 33.375

33.375 - 47.750

47.750 - 62.125

62.125 - 76.500

76.500 - 90.875

90.875 - 105.250

105.250 - 119.625

119.625 - 134.

The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.

The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

A similar problem is given at brainly.com/question/16631975

6 0

3 years ago

I need some assistance

motikmotik

I think B I'm not sure but if I think it's B, if it is not, let me know.

8 0

3 years ago

Find the exact length of the curve. x = 9 + 9t2, y = 6 + 6t3, 0 ≤ t ≤ 4

andrew11 [14]

ANSWER

$=102 \sqrt{ 17}$

EXPLANATION

The equation of the curve is:

$x = 9 + 9 {t}^{2}$

$y = 6 + 6 {t}^{3}$

We differentiate to obtain:

$\frac{dx}{dt} = 18t$

$\frac{dy}{dt} = 18t$

The length of the arc

$l = \int _0 ^{4} \sqrt{ {( \frac{dx}{dt} )}^{2} + {( \frac{dy}{dt} )}^{2} } dt$

This implies that:

$= \int _0 ^{4} \sqrt{ {( 18t )}^{2} + {( {18 {t}^{2} } )}^{2} } dt$

$= \int _0 ^{4} 18t \sqrt{ 1+ { {18 {t}^{2} } }} dt$

$=102 \sqrt{ 17}$

3 0

3 years ago

-10=x/2+6 Answers -8 -14 -32 18