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Sedbober [7]
3 years ago
14

Find the area of the given shape

Mathematics
1 answer:
mel-nik [20]3 years ago
6 0

Answer:

Here are some formulas and some info to get you started:

Step-by-step explanation:

Parallelogram:

Area = Base · Height

Triangle:

Area = 1/2 · Base · Height

Know that the height is the vertical dotted line. Also know that the horizontal dotted line attached to the base is not actually part of the base.

If you need any more help let me know, and good luck!

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How do i solve this radical expression? It has to do with factoring i know that.
MAVERICK [17]
Firstly it's not a radical expression.

You can factorize the numerator x²+12x+36 = (x+6)²

Factorize the denominator 12x+72 = 12(x+6)


=(x+6)² / 12(x+6) , simplify. Answer = (x+6)/12
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3 years ago
How to solve -3 1/2 + 5 4/5?
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Joan says that the expression
Gemiola [76]

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Explanation

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3 years ago
Complete each calculation using the distributive property.
Alisiya [41]
What are the answer options?
7 0
3 years ago
find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[1,2,-2]. B
SashulF [63]

Answer:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

Step-by-step explanation:

For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

a=[1,2,-2], b=[4,0,-3,]

The dot product on this case is:

a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

|a|= \sqrt{(1)^2 +(2)^2 +(-2)^2}=\sqrt{9} =3

|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5

And finally we can calculate the angle between the vectors like this:

cos \theta = \frac{ab}{|a| |b|}

And the angle is given by:

\theta = cos^{-1} (\frac{ab}{|a| |b|})

If we replace we got:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

3 0
3 years ago
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