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belka [17]
2 years ago
7

Suppose that a rectangular picture is surrounded by a frame of uniform width of x

Mathematics
1 answer:
user100 [1]2 years ago
3 0

Answer:

Ok so the area of the frame is 18*12 = 216cm^2

Let’s suppose the width of the frame is x

The total area (painting and frame) is (2x+18)(2x+12) and this is equal to 432

4x^2 + 60x + 216 = 432

4x^2 + 60x - 216 = 0

x^2 + 15x - 54 = 0

(x+18)(x-3) = 0

Therefore x = {-18,3}

Since width of the frame has to be positive, the width has to be 3cm Step-

i hope i helped!

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Solve the inequality 15 ≥ q + 3
oksano4ka [1.4K]

Answer:

q\le \:12

Step-by-step explanation:

<u>Solving</u>

\mathrm{Switch\:sides}  (Not a required step)

q+3\le \:15

\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}

q+3-3\le \:15-3\\=q\le \:12

\bold{q\le \:12}

<u>Graphing</u>

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2 years ago
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A group of market women sell at least one of yam, plantain and maize.12 of them sell maize,10 sell yam and 14 sell plantain,5 se
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Let the three items be M, Y and P.

n{M ∩ Y} only = 4-3 = 1

n{M ∩ P) only = 5-3 = 2

n{ Y ∩ P} only = 2

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3 years ago
In a class in which the final course grade depends entirely on the average of four equally weighted 100-point tests, Joyce has s
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Answer:

66 ≤ f ≤100

Explanation

Mean= ( Σ x ) / n

Mean= sum of scores/ number of subject she took

Now, she already too 3 subject which sum is 85+83+86=254

Now we need to know range of score for her to have (grade) a mark between 80 and 89

Now let take the lower limit mean=80

The lowest score she can get is

Mean = ( Σx) / n

80=(85+83+86+f)/4

80×4= 254+f

Therefore, f= 320-254=66

Therefore the minimum score she can have to have a B is 66.

Then, let take the upper limit mean 89. i.e the maximum she can have so that she don't have an A grade.

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Therefore this shows that she cannot have an A grade in the exam. The maximum score for the exam is 100.

There the range of score is 66 ≤ f ≤100 to have a B grade

66 ≤ f ≤100 answer

Since she cannot score 102 in the examination.

8 0
3 years ago
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