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notka56 [123]
3 years ago
6

The rectangle in the following figure has a base that equals three times its height. Find an expression for the rectangle's peri

meter (P) as a function of its base.

Mathematics
1 answer:
Tomtit [17]3 years ago
7 0

Ok. Length = L, Base = B

P = 2L + 2B

We know that B = 3L or L = B/3

so that's P = 2/3B + 2B = 2/3B + 6/3B = 8/3B

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Aleonysh [2.5K]
6x +5 = 8x after make the common denominator on both sides 

5 = 2x

x = 5/2 

hope this will help you 
5 0
3 years ago
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Which coordinate pairs make the equation x-9y=12 true?
asambeis [7]

Answer:

(12,0) and (3, -1) and (0,-4/3)

Step-by-step explanation:

To find if a point makes an equation true, just substitute its x and y values in in the equation and simplify.

1) Let's test the point (12,0). Substitute 12 for x and 0 for y in the equation:

x-9y=12\\(12)-9(0)=12\\12-0=12\\12=12

12 does equal 12, so (12,0) makes the equation true.

2) Let's test the point (0, 12). Substitute 0 for x and 12 for y in the equation:

x-9y=12\\(0)-9(12) = 12\\0-108 = 12\\-108 = 12

However, -108 does not equal 12, so (0,12) does not make the equation true.

3) Let's test the point (3, -1). Substitute 3 for x and -1 for y in the equation:

x-9y=12\\(3)-9(-1) = 12\\3 + 9 = 12\\12 = 12

12 does equal 12, so (3, -1) makes the equation true.

4) Let's test the point (0, -4/3). Substitute 0 for x and -4/3 for y in the equation:

(0) -9(-\frac{4}{3} ) = 12\\0 + \frac{36}{3} = 12\\0 + 12 = 12 \\12 = 12

12 does equal 12, so (0, -4/3) makes the equation true.

8 0
3 years ago
Please help me out with this
andre [41]

Answer:

\huge\boxed{\sqrt[3]{c^4}=c^\frac{4}{3}}

Step-by-step explanation:

\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\text{therefore}\\\\\sqrt[3]{c^4}=c^\frac{4}{3}

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3 years ago
What is the slope of the line described by y=3x-6
aleksandrvk [35]
Answer:

The Slope: m=3

6 0
3 years ago
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Solve the system of equations.
guajiro [1.7K]

Answer:

Option 2: (1, 0) and (0, -5)

Step-by-step explanation:

Let's solve this system of equations using the elimination method.

Start by labelling the two equations.

5x -y= 5 -----(1)

5x² -y= 5 -----(2)

(2) -(1):

5x² -y -(5x -y)= 5 -5

Expand:

5x² -y -5x +y= 0

5x² -5x= 0

Factorise:

5x(x -1)= 0

5x= 0 or x -1= 0

x= 0 or x= 1

Now that we have found the x values, we can substitute them into either equations to solve for y.

Substitute into (1):

5(0) -y= 5 or 5(1) -y= 5

0 -y= 5 or -y= 5 -5

y= -5 or -y= 0

y= 0

Thus, the solutions are (0, -5) and (1, 0).

3 0
2 years ago
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