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Zigmanuir [339]
3 years ago
9

Manny has 48 feet of wood.He wants to use all of it to create a border around a garden.The equation 2l+2w=48 can be used to find

the length and width of the garden, where l is the length and w is the width of the garden.If Manny makes the garden 15 feet long, how wide should the garden be?
Mathematics
1 answer:
schepotkina [342]3 years ago
4 0

Answer:

If Manny makes the garden 15 feet long, then the width of the garden would be 9 feet wide.

Step-by-step explanation:

We are given an equation: <em>2l + 2w = 48</em>. We can use this equation and the information given to us to find how wide Manny's garden should be if he makes the garden 15 feet long. We can plug in 15 for <em>l.</em>

2(15) + 2w = 48

Multiply 15 by 2.

30 + 2w = 48

Subtract 30 from 48.

2w = 18

Now, divide 2 from 18.

w = 9

So, the width of Manny's garden should be 9 feet wide.

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PLEASE HELP ASAP
Mice21 [21]

Answer:

2,100 people

Step-by-step explanation:

~Divide to find the number of people in fifths

3,500 / 5 = 700

~Multiply to find 2/5 of the total people

700 * 2 = 1,400

~Subtract to find the number of people against building the school.

3,500 - 1,400 = 2,100

Best of Luck!

3 0
3 years ago
A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.
Artist 52 [7]

so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

6 0
3 years ago
80% of Vanessa's final grade is based on the average score of her quizzes, which is 73.
Misha Larkins [42]
80% of quizzes + 20% of final = final grade

80% * 73 + 20% * x = 70

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8 0
3 years ago
What is the measure of angle x?<br> Enter your answer in the box.<br> x =<br> UwU PLZ help
MissTica
I think it might be 43° because since the line is going through the middle, it would be 180-47-90=43 (the 90 came from the right angle)

hope this helps! :)
7 0
3 years ago
Read 2 more answers
Find the number of ways of arranging the numbers
Doss [256]

First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).

So, we can think of our numbers as

\begin{array}{c|c}x&x\mod 3\\0&0\\1&1\\2&2\\3&0\\4&1\\5&2\\6&0\\7&1\\8&2\\9&0\end{array}

In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.

For example, we could arrange our numbers following the pattern

0,1,2,0,1,2,0,1,2

or

2,0,1,2,0,1,2,0,1

We have 3!=6 possible patterns. Suppose for example that we choose the pattern

0,1,2,0,1,2,0,1,2

One possible way of following this pattern would be the arrangement

3,1,2,6,4,5,9,7,8

In fact, we substituted every '0' with a multiple of 3 (3, 6 or 9), every '1' with a number 1 away from a multiple of 3 (1, 4 or 7) and every '2' with a number 2 away from a multiple of 3 (2, 5 or 8).

This means that, once we fix a patter, we have 3 choices for the first 3 slots, 2 choices for the next 3 slots, and the final slot will be fixed. So, we have

3\cdot 3\cdot 3\cdot 2 \cdot 2 \cdot 2 = 216

possible ways of following a fixed pattern. Since the number of patterns was 6, we have

216\cdot 6 = 1296

possible arrangements.

7 0
3 years ago
Read 2 more answers
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