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Neko [114]
4 years ago
5

Susan has a balance of $110 in her school lunch account, and she spends $6 per day. Her sister Evelyn has $83 in her lunch accou

nt, and she spends $4 per day for lunch. In how many days will Susan's lunch account balance be less than Evelyn's
Mathematics
1 answer:
Oxana [17]4 years ago
5 0

Answer:

In 2 days  Susan's lunch account balance will be less than Evelyn's

Step-by-step explanation:

Susan has a balance of $ 110  which will be enough for = 110/6= 18  days

Evelyn has a balance $ 83 which will be enough for = 83/4= 20 days

In 2 days  Susan's lunch account balance will be less than Evelyn's

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PLEASE HELP WILL GIVE BRAINLIEST<br> Answer all parts of the question please!
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A: l x w = a formula for area
l = w + 48 length
w(w+48) = 3024 substitute l with w+48
w^2 + 48w -3024=0 —solution to part A

B: w^2 + 48w - 3024 = 0 factor
(w-36)(w+84) = 0
Solutions for width = 36,-84
A width can’t be negative, so the width is 36 inches


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3 years ago
Find f if f′(x)=2x−3/x^4, x&gt;0 and f(1)=2.
slavikrds [6]
f'(x)=2x-\dfrac3{x^4}
\displaystyle\int f'(x)\,\mathrm dx=\int\left(2x-\frac3{x^4}\right)\,\mathrm dx
f(x)=x^2+\dfrac1{x^3}+C

Given that f(1)=2, we get

2=1^2+\dfrac1{1^3}+C
2=2+C
\implies C=2

so that

f(x)=x^2+\dfrac1{x^3}
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You guessed the height of the building to be 23 feet, but it was actually 35 feet. What was your percent error? Round to the nea
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The percent error is 34.28 %.
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8 0
3 years ago
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Square root of 9 multiplied by square root of 180
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In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.
konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

BC^2=BD\cdot AB.

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.

Take positive value x. You get

AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.

2. According to the previous theorem,

AC^2=AD\cdot AB.

Then

AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

Answer: AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D

This means that you cannot find solutions of this equation. Then CD≠2 cm.

8 0
4 years ago
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