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Eduardwww [97]
3 years ago
15

What is the area of rectangle ABCD in square units? , I’ll mark brainliest

Mathematics
1 answer:
tiny-mole [99]3 years ago
3 0

Answer:

What is the area of rectangle ABCD in square units?

, I’ll mark brainliest

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Write a Function rule for the table.
Brut [27]

f(x) = x + 3

when x = 0, f(x) = 3

when x = 1, f(x) = 1 + 3 = 4

when x = 2, f(x) = 2 + 3 = 5

when x = 3, f(x) = 3 + 3 = 6

answer is D

f(x) = x + 3

6 0
3 years ago
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1+4=5<br> 2+5=12<br> 3+6=21<br> 8+11=?
Mrac [35]

Answer:40

Step-by-step explanation:

21+8=29

29+11=40

6 0
3 years ago
Which shape has no parallel sides?<br> rectangle<br> trapezoid<br> square<br> triangle
nadezda [96]

Answer:

Triangle

Step-by-step explanation:

all of them have parallel sides except triangle

5 0
2 years ago
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Write the quadratic equation that has roots -1-rt2/3 and -1+rt2/3 if its coefficient with x^2 is equal to 1
weeeeeb [17]

The equation of the quadratic function is f(x) = x²+ 2/3x - 1/9

<h3>How to determine the quadratic equation?</h3>

From the question, the given parameters are:

Roots = (-1 - √2)/3 and (-1 + √2)/3

The quadratic equation is then calculated as

f(x) = The products of (x - roots)

Substitute the known values in the above equation

So, we have the following equation

f(x) = (x - \frac{-1-\sqrt{2}}{3})(x - \frac{-1+\sqrt{2}}{3})

This gives

f(x) = (x + \frac{1+\sqrt{2}}{3})(x + \frac{1-\sqrt{2}}{3})

Evaluate the products

f(x) = (x^2 + \frac{1+\sqrt{2}}{3}x + \frac{1-\sqrt{2}}{3}x + (\frac{1-\sqrt{2}}{3})(\frac{1+\sqrt{2}}{3})

Evaluate the like terms

f(x) = x^2 + \frac{2}{3}x - \frac{1}{9}

So, we have

f(x) = x²+ 2/3x - 1/9

Read more about quadratic equations at

brainly.com/question/1214333

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7 0
1 year ago
Find the y-intercept of the line: -4x - 2y = 16
Georgia [21]
The y-intercept is (0, -8) 
if you set the equation to slope intercept format you get

y=-2x-8
3 0
3 years ago
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