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marta [7]
4 years ago
6

A teacher noticed 5/8 of the students were wearing either blue shorts or white shorts. Write two different ways this could be do

ne
Mathematics
1 answer:
lukranit [14]4 years ago
3 0
3/8 could be wearing blue shorts and 2/8 could be wearing white shorts
4/8 could be wearing white shorts and 1/8 could be wearing blue shorts
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The perimeter of a rectangle must be no greater than 62 meters. The width must be 12
GREYUIT [131]

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xinloitoikogiaiduoc

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The answer is 8 units
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Which inequality matches the graph?
alexdok [17]

The inequality that matches the graph would be C. -3x + 2y  ≥ 7.

4 0
3 years ago
Casey bought a 15.4 pound turkey and an 11.6 pound ham for Thanksgiving and paid $38.51.
White raven [17]

Answer:

h =  1.74

t =  1.19

Step-by-step explanation:

15.4t + 11.6h = 38.51

10.2t + 7.3h = 24.84

15.4•t + 11.6•h = 38.51

10.2•t + 7.3•h = 24.84

11.6h+15.4t = 38.51

7.3h+10.2t = 24.84

h = 87

50

= 1.74

t = 119

100

= 1.19

6 0
3 years ago
Can anyone explain to me how to do this question ? ty !!​
julia-pushkina [17]

Step-by-step explanation:

use a little imagination.

the whole thing is a combination of churches and half- circles.

and you know the formula for the area of a circle ?

A = pi×r²

so, the only difficulty is to find the circles and then the radius of each involved circle.

when you look at the picture, the shaded area is almost identical to the lower half of the main circle.

2 bumps "look" over into the other side, but one equally sized bump is removed from this lower half.

so, as a first step we will calculate the area of the lower half of the main circle.

the diameter is 24 m, so the radius is half of that : 12m

the area of the large circle is therefore

pi×12² = 144×pi m²

since we only need the (lower) half, we divide this by 2 :

144×pi / 2 = 72×pi m²

the bumps up and down are actually half-circles too. just of smaller circles.

since AB = BC = CD, we know each of these lines are 1/3 of the main diameter AD.

so, the diameter of the smaller circles is 24/3 = 8 m

and their radius is half of that = 8/2 = 4m

the area of such a small circle is therefore

As = pi×4² = 16×pi m²

and a small half-circle area is then

16×pi / 2 = 8×pi m²

so, for the full shaded area I need to take the large half-circle and add 2 small half-circles, and then subtract 1 small half-circle.

so, in total I only need to add 1 small half-circle.

Ashaded = 72×pi + 8×pi = 80×pi m²

5 0
3 years ago
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