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julsineya [31]
3 years ago
9

Micah created a border for his school photo. He cut a 54 cm ribbon to go around the photo. His photo measures 12 cm on one side.

What is the area of his photo?
Mathematics
1 answer:
Paha777 [63]3 years ago
6 0

Answer:

648

Step-by-step explanation:

54X12=648 cm squared

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Which statement is true
Eva8 [605]

Answer:

3/6 = 4/8

Step-by-step explanation:

they're both halves

8 0
3 years ago
Read 2 more answers
I Need help! If you help me you'll get 10 points!!
natka813 [3]

To solve this problem you must apply the proccedure shown below:

1. You have that the number of innings are expressed as a mixed number in the exercise:

102^{\frac{2}{3} }

2. If you want to write it as a decimal, you can convert the fraction as a decimal by dividing the numerator by the denominator and then, you must add this to the whole number part:

102+\frac{2}{3}=102+0.67=102.67

The answer is: 102.67

7 0
3 years ago
Problem solving your family has a rectangular pool that measures 18 feet by 9 feet. your family wants to put a deck around the p
tester [92]
Let w be the width of the deck. Then, using the total area, we get:
(9+2w)(18+2w)=400
4w²+54w-238=0
2w²+27w-119=0
(2w-7)(w+17)=0
w=-17 or 7/2
Using the positive value, we get a width of 3&1/2 ft. for the deck
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7 0
3 years ago
The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t
Doss [256]

Answer:

First Case:

\displaystyle p=\frac{5}{2}\text{ and } d=-2

Second Case:

\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

6p+2, 4p^2-10, \text{ and } 4p+3

Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.

Therefore, we can write the second term as;

4p^2-10=(6p+2)+d

And, likewise, for the third term:

4p+3=(6p+2)+2d

Let's solve for <em>d</em> for each of the equations.

Subtracting in the first equation yields:

d=4p^2-6p-12

And for the second equation:

2d=-2p+1

To avoid fractions, let's multiply the first equation by 2. Hence:

2d=8p^2-12p-24

Therefore:

8p^2-12p-24=-2p+1

Simplifying yields:

8p^2-10p-25=0

Solve for <em>p</em>. We can factor:

8p^2+10p-20p-25=0

Factor:

2p(4p+5)-5(4p+5)=0

Grouping:

(2p-5)(4p+5)=0

Zero Product Property:

\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}

Then, we can use the second equation to solve for <em>d</em>. So:

2d_1=-2p_1+1

Substituting:

\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}

So, for the first case, <em>p</em> is 5/2 and <em>d</em> is -2.

Likewise, for the second case:

\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}

So, for the second case, <em>p </em>is -5/4, and <em>d</em> is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

For Case 2, we will have:

-11/2, -15/4, -2.

8 0
2 years ago
Solve the system of equations 3r – 4s = 0 and 2r 5s = 23.
UkoKoshka [18]
this is when its solved: r=4, s=3}
3 0
3 years ago
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