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sdas [7]
3 years ago
11

Find the area of the parallelogram

Mathematics
2 answers:
dimulka [17.4K]3 years ago
8 0

Answer:

6 cm squared

Step-by-step explanation:

Parallelogram: Base*Height=Area

Base: 3

Height: 2

3*2=6 cm squared

ryzh [129]3 years ago
6 0

Answer:

8

Step-by-step explanation:

Its 8 because when you find the area it is just Base times height so when you do that you will get 8 for your answer.

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mars1129 [50]

Answer:

x = 76/75

Step-by-step explanation:

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3 years ago
A Converse Statement is when you turn the statement into a double negative?
Misha Larkins [42]

Answer:

False

Step-by-step explanation:

This is because there is no such thing! A double negative becomes a positive! Plz give brainliest!

3 0
3 years ago
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What single percentage change is equivalent to a 11% increase followed by a 15% increase?​
AlekseyPX

Answer:

It's a total increase by 27,65%

Step-by-step explanation:

1,11 × 1,15 = 1,2765

1,2765 = +27,65%

7 0
3 years ago
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lozanna [386]

Answer:

Answer in explanation

Step-by-step explanation:

Before this we do this question let's get fimiliar with what's a function.

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The vertical line test is just puting a vertical line anywhere on the function. If the vertical line touches more than two points your graph is not a function. If it touches one you graph is a function. The reason is because a vertical line that touches 2 points means one x maps to two diffrent y and thats not a function.

In conclusion the graph does pass the vertical line test so it is a function.

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JoeLouis2

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3 years ago
Given a triangle JLM with vertices J(-4,3), L(-2,7), and M(2,7). What would the coordinates be if it was reflected across the li
topjm [15]

Answer: M'(2, - 5), L'(-2, -5), j'(-4, - 1)

Step-by-step explanation:

When we do a reflection over a given line, the distance between all the points (measured perpendicularly to the line) does not change.

The line is y = 1.

Notice that a reflection over a line y = a (for any real value a) only changes the value of the variable y.

Let's reflect the points:

J(-4, 3)

The distance between 3 and 1 is:

D = 3 - 1 = 2.

Then the new value of y must also be at a distance 2 of the line y = 1

1 - 2 = 1

The new point is:

j'(-4, - 1)

L(-2, 7)

The distance between 7 and 1 is:

7 - 1 = 6.

The new value of y will be:

1 - 6 = -5

The new point is:

L'(-2, -5)

M(2,7)

Same as above, the new point will be:

M'(2, - 5)

3 0
3 years ago
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