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3 years ago

Help Me Number 6 or if you can help me with number 7 if you want too ☺️

Mathematics

1 answer:

Makovka662 [10]3 years ago

7 0

Simply use the formula they give you. For question 6, 4 is the radius. So the area of the circle is 4pi. Now for seven you do the same thing, but they gave you the diameter instead and it’s a half circle. So try that one on your own. Let me know if you get stuck

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A function f(x) is graphed. What is the slope of the function? m= What is the y-intercept of the function? b= Which equation rep

timofeeve [1]

Answer:

slope is m = 2

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Drag the tiles to list the sides of △MNO from shortest to longest.

sweet [91]

The smaller the angle subtended by a side, the smaller the length of the

side.

The correct responses are;

Question 1: The list of sides from shortest to longest are;

MO/Shortest MO/Medium and MO/Longest

a) Friday

b) 70 minutes

c) 40%

d) Yes, the sum of the mean number of minutes spent on aerobic training and the mean number of minutes spent on strength training is equal to the mean total number of minutes spent training.

From the given diagram, we have, the measure of the third angle, ∠O, is

found as follows;

∠O = 180° - 54° - 61° = 65°

Therefore, ∠O = The largest angle

We get;

The longest side is opposite the largest angle, which gives;

The shortest side is the side opposite ∠N (54°)= $\frac{}{MO}$

The next shortest side is the side opposite ∠M(61°) = $\frac{}{NO}$

The longest side is the side opposite ∠O(65°) = $\frac{}{MN}$

a) The time spent training on Tuesday = 60 + 10 = 70 minutes

The time spent training on Thursday = 50 + 30 = 80 minutes

The time spent training on Friday = 45 + 40 = 85 minutes

Therefore, the day the athlete spent the longest total amount of time training is on Friday

b) The time spent training on Monday = 10 + 20 = 30 minutes

The time spent training on Wednesday = 20 + 15 = 35 minutes

Therefore, we get;

30, 35, 70, 80, and 85

The median total number of minutes the athlete spent training each day = 70 minutes

c) The time spent strength training = 20 + 10 + 15 + 30 + 45 = 120

The total number of minutes the athlete spent training = 70 + 80 + 85 + 30 + 35 = 300

$The$ $percentage$ $spent$ $on$ $strength$ $training$ = $\frac{120}{300}$ × $100 =$ $\frac{40}$ %

d) The mean number of minutes spent on strength training is found as follows;

$Mean_{strength} =\frac{120}{5} =24$

The mean number of minutes spent on aerobic training is found as follows;

$Mean_{aerobic} =\frac{10+60+20+50+40}{5} =36$

$Mean_{strength} +Mean_{aerobic} =24+36=60$

The mean total number of minutes spent training, $Mean_{total}$ = $\frac{300}{5}$ = 60

Therefore;

$Mean_{strength}+Mean_{aerobic} = Mean_{total} \\$

Learn more here:

brainly.com/question/2962546

4 0

2 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

Step-by-step explanation:

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$