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Vladimir79 [104]

3 years ago

11

I WILL GIVE BRAINLIEST IF IT IS CORRECT! The Atlantic Ocean covers about 3.2 × 10^7 square miles. The Pacific Ocean covers about

6.4 × 10^7 square miles. About how much area does the Atlantic Ocean and Pacific Ocean cover? A. 3.2 × 10^7 square miles B. 9.6 × 10^7 square miles C. 3.2 × 10^14 square miles D. 9.6 × 10^14 square miles

2 answers:

kkurt [141]3 years ago

3 0

Answer:

B

Step-by-step explanation:

3.2 * 10⁷ + 6.4 * 10⁷

= 10⁷ * (3.2 + 6.4)

= 10⁷ * 9.6

= 9.6 * 10⁷

Ghella [55]3 years ago

3 0

Answer:

The Atlantic Ocean and Pacific Ocean covers about 9.6 x 10^7 mi^2

Step-by-step explanation:

Let

x -----> area covered by the Atlantic Ocean

y -----> area covered by the Pacific Ocean

we know that

The area that covers the Atlantic Ocean and Pacific Ocean is equal to the sum of the area covered by the Atlantic Ocean and the area covered by the Pacific Ocean

x = 3.2 x 10^7

y = 6.4 x 10^7

Substitute

x + y = 3.2 x 10^7 + 6.4 x 10^7 = 9.6 x 10^7.

Hope this helps! And good luck on your test!

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Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

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x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

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