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Alex777 [14]
2 years ago
5

a banner is made of a sqiare and a semocircle. the square has side lengths of 26 inches. one side of the square is also the diam

eter of the semicircle. what is the total area of the banner?
Mathematics
2 answers:
tamaranim1 [39]2 years ago
8 0
ARea = area of square + area of semicircle

         = 26^2 + 0.5pi(13)^2

        = 941.46  square inches  to nearest hundredth
denpristay [2]2 years ago
6 0
To find the area of the square, multiply the length and height(676 sq in). To find the semicircle square the radius (half the diameter) and multiply by 3.14 (or pi) Then divide by two. Giving you a total area of 941.33 sq inches.
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What is 6x – 8 = 4 written as a system? y = 6x – 8 y = –4 y = 6x – 8 y = 4 y = –6x + 8 y = 4
mestny [16]

6x – 8 = 4 written as a system. Hence second option is correct

<u>Solution:</u>

Given, equation is 6x – 8 = 4.

We have to find the equivalent system from given options.

First let us solve given equation, 6x – 8 = 4 ⇒ 6x = 4 + 8 ⇒ 6x = 12 ⇒ x = 2

Now, let us solve one by one option ,so that when we get above solution, that is correct option.

  • y = 6x – 8 and y = -4 ⇒ put y = -4 in y = 6x – 8 ⇒ 6x – 8 = -4 ⇒ 6x = 8 – 4

⇒ 6x = 4 ⇒ x ≠ 2, wrong option

  • y = 6x – 8 and y = 4 ⇒ put y = 4 in y = 6x – 8 ⇒ 6x – 8 = 4 ⇒ 6x = 8 + 4 ⇒ 6x = 12 ⇒ x = 2, right option.
  • y = -6x + 8 and y = 4 ⇒ put y = 4 in y = -6x + 8 ⇒ -6x + 8 = 4 ⇒ 6x = 8 – 4 ⇒ 6x = 4 ⇒ x ≠ 2, wrong option

hence, second option is correct.

6 0
3 years ago
Read 2 more answers
Hi how can I graph this if it dosn't have a slope?<br> 1. y = (2/3)x - 1 <br> y = -x + 4
12345 [234]

System of Linear Equations entered :

[1] y - 2x/3 = -1

[2] y + x = 4

// To remove fractions, multiply equations by their respective LCD

Multiply equation [1] by 3

// Equations now take the shape:

[1] 3y - 2x = -3

[2] y + x = 4

Graphic Representation of the Equations :

-2x + 3y = -3 x + y = 4

Solve by Substitution :

// Solve equation [2] for the variable x

[2] x = -y + 4

// Plug this in for variable x in equation [1]

[1] 3y - 2•(-y +4) = -3

[1] 5y = 5

// Solve equation [1] for the variable y

[1] 5y = 5

[1] y = 1

// By now we know this much :

y = 1

x = -y+4

// Use the y value to solve for x

x = -(1)+4 = 3

I hope this help you

8 0
3 years ago
Please please please helppppppppp
notka56 [123]

Answer:

Step-by-step explanation:

She added 3x and -2x, getting 5x, instead of x. She needs to pay more attention to the positive/negative signs.

6 0
2 years ago
1. A function has a second rate of change of -4. Does the function have a maximum or a minimum?
Aloiza [94]

This function would have a maximum.

Since we are subtracting by a -4 for each increase in x, we know that the numbers will continue to go down. Given this fact, we know the number will never be higher than when we started, but the number could go infinitely low. As a result we have a maximum and no minimum.

8 0
2 years ago
Solve.
Nikitich [7]
The answer is C: 10\text{ }^1/_2. Here are the details:

\text{Equation:}\\ 6\text{ }^1/_3+10\text{ }^1/_2\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{Start with the integers.}\\ 6+10=16\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{...then with the fractions, but rewrite them first to make it easier.}\\ ^1/_3+\text{ }^1/_2\stackrel{\text{rewrite}}{\to}\text{ }^2/_6+\text{ }^3/_6=\text{ }^5/_6\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{Add 'em up!}\\&#10;16\text{ }^5/_6

\text{Equation:}\\&#10;16\text{ }^5/_6+3\text{ }^5/_6\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{Start with the integers.}\\&#10;16+3=19\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{...then with the fractions, but rewrite them first to make it easier.}\\&#10;^5/_6+\text{ }^5/_6=\text{ }^{10}/_6\stackrel{\text{rewrite}}{\to}\text{ }^5/_3\stackrel{\text{rewrite}}{\to}1\text{ }^2/_3\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{Add 'em up!}\\&#10;20\text{ }^2/_3

\text{Last equation:}\\ 20\text{ }^2/_3+5\text{ }^1/_2\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{Start with the integers.}\\ 20+5=25\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{...then with the fractions, but rewrite them first to make it easier.}\\ ^2/_3+\text{ }^1/_2\stackrel{\text{rewrite}}{\to}\text{ }^4/_6+\text{ }^3/_6=\text{ }^7/_6\stackrel{\text{rewrite}}{\to}1\text{ }^1/_6\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{Add the integer and fraction together.}\\&#10;25+1\text{ }^1/_6\stackrel{?}{=}26\text{ }^1/_6

26\text{ }^1/_6\stackrel{\checkmark}{=}26\text{ }^1/_6
5 0
2 years ago
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