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3 years ago

8

Someone helppppp mee please !

1 answer:

Alchen [17]3 years ago

4 0

<span>Area = (1/2)(533 ft)(525 ft) sin(53) = 111739 square feet</span>

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I will mark brainliest please help

amid [387]

DGE, it’s a vertical angle

4 0

3 years ago

A ship traveled for a total of 147 miles over the course of 8 hours. Heading south, the ship traveled at an average speed of 15

Cloud [144]

Alright

s=number of hours headd south
e=number of hours headed east

s+e=8

and
total is 147 miles
distance=speed times time
15s+24e=147

we got 2 equations

s+e=8
15s+24e=147
subsitution

s+e=8
minus e both sides
s=8-e
subsitut 8-e for s in other equaton

15s+24e=147
15(8-e)+24e=147
120-15e+24e=147
120+9e=147
minus 120 both sides
9e=27
divide both isdes by 9
e=3

sub back
s=8-e
s=8-3
s=5

it was headed south 5 miles

6 0

3 years ago

Of 400 votes cast, olivia received 38% of the votes and michael received 116 votes. what percentage did michael receive?

ch4aika [34]

Of 400 votes cast, olivia received 38% of the votes and michael received 116 votes. what percentage

6 0

3 years ago

Find the area of the shaded region in square units. Show your reasoning.

motikmotik

Answer:

40 square units

Step-by-step explanation:

First of all, lets say that square has side $l$ , so, the area unit is $l^2$

the diagonal's square is $l\sqrt{2}$

CALCULATION OF TRIANGLES'S AREA (there are 4 triangles)

$A_{triangles}=4*base*heigh*0.5=2*(l\sqrt{2} )(2l\sqrt{2} )=8l^2$

CALCULATION OF MAIN SQUARE AREA

$A_{square}=side*side=(4\sqrt{2} l)(4\sqrt{2} l)=32l^2$

TOTAL AREA

$A_{total}=A_{triangles}+A_{square}=8l^2+32l^2=40l^2$

3 0

3 years ago

Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.

GREYUIT [131]

Answer:

$y=\dfrac{1}{1-Ke^{-t}}$

Step-by-step explanation:

Given

The given equation is a differential equation

$\dfrac{dy}{dt}=y-y^2$

$\dfrac{dy}{dt}=-(y^2-y)$

By separating variable

⇒ $\dfrac{dy}{(y^2-y)}=-t$

$\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt$

Now by taking integration both side

$\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt$

⇒ $\ln (y-1)-\ln y=-t+C$

Where C is the constant

$\ln \dfrac{y-1}{y}=-t+C$

$\dfrac{y-1}{y}=e^{-t+c}$

$\dfrac{y-1}{y}=Ke^{-t}$

$y=\dfrac{1}{1-Ke^{-t}}$

from above equation we can say that

When t will increases in positive direction then $e^{-t}$ will decreases it means that ${1-Ke^{-t}}$ will increases, so y will decreases. Similarly in the case of negative t.

4 0

3 years ago