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4 years ago

Please show me how you got the answer. thanks in advance

Mathematics

1 answer:

lesantik [10]4 years ago

5 0

Write and solve an equation of ratios:

6 inches 1 inch
-------------- = ------------
60 inches x

Cross-multiplying, 6x = 60, so x = 10 inches (answer)

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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

3 years ago

Alaina evaluated 17/18 + 1/20 and got an answer of 18/38. which statement is true about her answer?

Basile [38]

Um can i have answer choices please? the. i’ll answer the question

3 0

3 years ago

Solve each system by substitution. y=3x+4 y=4x+7

Y_Kistochka [10]

We need to solve the following system by substituition:

$\begin{cases}y=3x+4 \\ y=4x+7\end{cases}$

To solve by substituition means that we need to isolate one variable in one of the equations and replace one expression with the other. This is done below:

$\begin{gathered} (3x+4)=4x+7 \\ 3x+4=4x+7 \\ 3x-4x=7-4 \\ -x=3 \\ x=-3 \end{gathered}$

Then we need to replace the value of "x" in any of the two equations.

$\begin{gathered} y=3\cdot(-3)+4 \\ y=-9+4 \\ y=-5 \end{gathered}$

The solution is (-3,-5).

4 0

1 year ago

$$ 30 POINTS$$ PLEASE HELP!!!!

fomenos

60°, 120° and 180° are the angles at which it will map onto itself.

4 0

3 years ago

Read 2 more answers

Find the radius of a cone with volume 45.19 cm^3 and height 7.5 cm. Show work.

dybincka [34]

Radius of cone is2.39cmor 2.4cm volume of cone=45.19cm^3 Pi×r^2×h/3=45.19 22×r^2×7.5/7=45.19×3 r^2=45.19×3×7/22×7.5 r^2=948.99/165 r^2=5.7514545 r=2.39 Or we can say r=2.4cm

7 0

3 years ago