Mathematics, 04.12.2019 04:31 loganharper992

i have no idea how to do it
do all or at least a few and show
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i have no idea how to do it do all or at least a few and show will mark

Answers: 3

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Answer from: kat122402

Part 1) $c=\sqrt{146}\ cm$

Part 2) $c=8\sqrt{2}\ in$

Part 3) The diagonal of rectangle is $d=3\sqrt{29}\ cm$

Part 4) The length of the diagonal of computer monitor is $d=15\ in$

Part 5) The ramp is $10.59\ ft$ long

Part 6) The distance between their houses is $=4\sqrt{2}\ mi$

Part 7) $234.31\ mi$

Part 8) 120 feet of wire is required

Step-by-step explanation:

Part 1) we know that

To find the length of the hypotenuse in a right triangle apply the Pythagorean Theorem

c^2=a^2+b^2

where

c is the hypotenuse (the greater side)

a and b are the legs

we have

$a=11\ cm\\b=5\ cm$

substitute

c^2=11^2+5^2

c^2=146

$c=\sqrt{146}\ cm$

Part 2) we know that

To find the length of the hypotenuse in a right triangle apply the Pythagorean Theorem

c^2=a^2+b^2

where

c is the hypotenuse (the greater side)

a and b are the legs

we have

$a=8\ in\\b=8\ in$

substitute

c^2=8^2+8^2

c^2=128

$c=\sqrt{128}\ in$

simplify

$c=8\sqrt{2}\ in$

Part 3) we know that

To find the length of the diagonal in a rectangle apply the Pythagorean Theorem

d^2=b^2+h^2

where

d is the diagonal of rectangle

b and h are the base and the height of rectangle

we have

$b=15\ cm\\hb=6\ cm$

substitute

d^2=15^2+6^2

d^2=261

$d=\sqrt{261}\ cm$

simplify

$d=3\sqrt{29}\ cm$

Part 4) we know that

To find the length of the diagonal of a computer monitor apply the Pythagorean Theorem

d^2=w^2+h^2

where

d is the diagonal of computer monitor

w and h are the wide and the high of computer monitor

we have

$w=12\ in\\h=9\ in$

substitute

d^2=12^2+9^2

d^2=225

$d=\sqrt{225}\ in$

simplify

$d=15\ in$

Part 5) we know that

To find out the length of the ramp apply the Pythagorean Theorem

L^2=x^2+y^2

where

L is the length of the ramp

x is the horizontal distance of the ramp

y is the vertical distance of the ramp

we have

$x=10\ ft\\y=3.5\ ft$

substitute

L^2=10^2+3.5y^2

L^2=112.25

$L=10.59\ ft$

Part 6) we know that

To find the distance between their houses apply the Pythagorean Theorem

c^2=a^2+b^2

where

c is the hypotenuse (distance between their houses)

a and b are the legs

we have

$a=4\ mi\\b=4\ mi$

substitute

c^2=4^2+4^2

c^2=32

$c=4\sqrt{2}\ mi$

Part 7) we know that

The speed is equal to divide the distance by the time

speed=distance/time

the distance is equal to multiply the speed by the time

distance=speed*time

First train

speed=60 mph

time=3 hours

distance=60(3)=180 miles

Second train

speed=50 mph

time=3 hours

distance=50(3)=150 miles

To find out how far apart are the trains at the end of 3 hours apply the Pythagorean Theorem

c^2=a^2+b^2

where

c is the hypotenuse (distance between the trains)

a and b are the legs

we have

$a=180\ mi\\b=150\ mi$

substitute

c^2=180^2+150^2

c^2=54,900

$c=234.31\ mi$

Part 8) we know that

For one tree is needed three wire

To find out the length of one wire apply the Pythagorean Theorem

c^2=a^2+b^2

where

c is the hypotenuse (length of one wire)

a and b are the distance above the ground and the distance from the base

we have

$a=3\ ft\\b=4\ ft$

substitute

c^2=3^2+4^2

c^2=25

$c=5\ ft$

For one tree is required ----> (5)3=15 ft

therefore

For 8 trees is required

Multiply by 8

15(8)=120 ft

Answer from: Quest

step-by-step explanation:

Answer from: Quest

Answers4.) -1/205.) -1/406.) -55/127.) -3/4
Answer these i don’t know at all and need fast

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