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3 years ago

What is the area of the composite figure whose vertices have the following coordinates?

Mathematics

2 answers:

Fofino [41]3 years ago

8 0

Answer:

60 square units

Step-by-step explanation:

We can use Pythagoras to solve this as appropriately projecting lines from any point will form a triangle. The vertical and horizontal lengths of which can be read of from the axis.

The distance between two points is √ ( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2

Let the distance between two points be

d i

Distance between points A and B

d 1 = √ [ 3 − ( − 3 ) ] 2 + [ 7 − 5 ] 2 d 1 = √ 36 + 4 = √ 2 2 × 10 = 2 √ 10

Distance between points B and C

d2 = √ [ 6 − 3 ] 2 + [ 7 − ( − 2 ) ] 2 d 2 = √ 9+ 81 = √ 3 2 × 10 = 3 √ 10

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Area = d 1 × d 2 = 2 √ 10 × 3 √ 10 = 60

YW :>

AveGali [126]3 years ago

6 0

Answer:

60 square units

Step-by-step explanation:

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Read 2 more answers

The equation 2x^2 – 8x = 5 is rewritten in the form of 2(x – p)^2 + q = 0. what is the value of q?

QveST [7]

Answer:

q = -13

Step-by-step explanation:

The given equation is:

$2x^{2}-8x=5$

Taking 2 as common from left hand side, we get:

$2(x^{2}-4x)=5\\\\2[x^{2} - 2(x)(2)]=5$

The square of difference is written as:

$(a-b)^{2}=a^2 - 2ab + b^{2}$ Equation 1

If we compare the given equation from previous step to formula in Equation 1, we note that we have square of first term(x), twice the product of 1st term(x) and second term(2) and the square of second term(2) is missing. So in order to complete the square we need to add and subtract square of 2 to right hand side. i.e.

$2[x^{2}-2(x)(2)+(2)^2-(2)^{2}]=5\\\\ 2[x^{2}-2(x)(2)+(2)^2]-2(2)^{2}=5\\\\ 2(x-2)^{2}-2(4)=5\\\\ 2(x-2)^2-8=5\\\\ 2(x-2)^{2}-8-5=0\\\\ 2(x-2)^{2}-13=0$

Comparing the above equation with the given equation:

$2(x-p)^{2}+q=0$ , we can say:

p = 2 and q= -13

6 0

3 years ago