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Colt1911 [192]
3 years ago
5

Alan solved the proportion StartFraction x over 200 EndFraction = StartFraction 8 over 25 EndFraction as shown. StartFraction x

over 200 EndFraction = StartFraction 8 over 25 EndFraction. (8) (x) = (25) (200). 8 x = 5,000. StartFraction 8 x over 8 EndFraction = StartFraction 5,000 over 8 EndFraction. X = 625. What is Alan’s error? He got the wrong product when he multiplied 25 by 200. He got the wrong quotient when he divided 5,000 by 8. He mixed up the positions of 8 and 25 in the equation (8) (x) = (25) (200). He mixed up the positions of 8 and 200 in the equation (8) (x) = (25) (200)
Mathematics
2 answers:
dolphi86 [110]3 years ago
8 0

Answer:

C on edge

Step-by-step explanation:

Alan solved the proportion StartFraction x over 200 EndFraction = StartFraction 8 over 25 EndFraction as shown.

StartFraction x over 200 EndFraction = StartFraction 8 over 25 EndFraction. (8) (x) = (25) (200). 8 x = 5,000. StartFraction 8 x over 8 EndFraction = StartFraction 5,000 over 8 EndFraction. X = 625.

bulgar [2K]3 years ago
7 0

Answer:

Step-by-step explanation:

The proportion that Alan solved was

x/200 = 8/25

His working as shown was

(8)(x) = (25) (200)

8x = 5000

He divided both sides of the equation by 8. It became

8x/8 = 5000/8

x = 625

The correct steps are

25x = 200 × 8 = 1600

Dividing both sides of the equation by 25, it becomes

x = 1600/25

x = 64

Alan's error were:

1) He got the wrong product when he multiplied 25 by 200.

2)He got the wrong quotient when he divided 5,000 by 8.

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Answer:

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Step-by-step explanation:

hope this helps!!

3 0
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2.5

Step-by-step explanation:

7 divided by 28= 0.25 x 10 = 2.5

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\bf \begin{array}{ccllll}
amount&\$\\
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3 years ago
Find the value or measure. Assume all lines that appear to be tangent are tangent. X=
aev [14]

According to the secant-tangent theorem, we have the following expression:

(x+3)^2=10.8(19.2+10.8)

Now, we solve for <em>x</em>.

\begin{gathered} x^2+6x+9=10.8(30) \\ x^2+6x+9=324 \\ x^2+6x+9-324=0 \\ x^2+6x-315=0 \end{gathered}

Then, we use the quadratic formula:

x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

Where a = 1, b = 6, and c = -315.

\begin{gathered} x_{1,2}=\frac{-6\pm\sqrt[]{6^2-4\cdot1\cdot(-315)}}{2\cdot1} \\ x_{1,2}=\frac{-6\pm\sqrt[]{36+1260}}{2}=\frac{-6\pm\sqrt[]{1296}}{2} \\ x_{1,2}=\frac{-6\pm36}{2} \\ x_1=\frac{-6+36}{2}=\frac{30}{2}=15 \\ x_2=\frac{-6-36}{2}=\frac{-42}{2}=-21 \end{gathered}<h2>Hence, the answer is 15 because lengths can't be negative.</h2>
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1 year ago
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makvit [3.9K]

Answer:

At most, Liam can only buy lunch for 6 people.

Step-by-step explanation:

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All you have to do is divide 50 by 8 and round down because you don't want to spend more than 50 dollars.

Writing it mathematically, it would be:

p \leq 6

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