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weqwewe [10]
3 years ago
11

Find the midpoint of AC.

Mathematics
1 answer:
Bogdan [553]3 years ago
8 0

Answer:

(0+a)/2 , (0+a)/2

= (a/2, a/2)

Answered by GAUTHMATH

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The area of a certain desert desert one is five times the area of another desert desert 2 if the sum of their areas is 24, 000,
sweet-ann [11.9K]

4,000,000 and 20,000,000 miles²

let x be the area of one desert then 5x is the area of the other

sum = x + 5x = 24,000,000

6x = 24,000,000

divide both sides by 6

x = 4,000,000

one desert is 4,000,000 miles²

and the other is (5 × 4,000,000)= 20,000,000 miles²


8 0
4 years ago
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Find the value of kk for which the constant function x(t)=kx(t)=k is a solution of the differential equation 5t3dxdt+2x−2=05t3dx
uranmaximum [27]
Given the differential equation

5t^3 \frac{dx}{dt} +2x-2=0

The solution is as follows:

5t^3 \frac{dx}{dt} +2x-2=0 \\  \\ \Rightarrow5t^3 \frac{dx}{dt} =2-2x \\  \\ \Rightarrow \frac{5}{2-2x} dx= \frac{1}{t^3} dt \\  \\ \Rightarrow \int {\frac{5}{2-2x} } \, dx = \int {\frac{1}{t^3}} \, dt \\  \\ \Rightarrow- \frac{5}{2} \ln(2-2x)=- \frac{1}{2t^2} +A \\  \\ \Rightarrow\ln(2-2x)= \frac{1}{5t^2} +B\\  \\ \Rightarrow2-2x=Ce^{\frac{1}{5t^2}} \\  \\ \Rightarrow 2x=Ce^{\frac{1}{5t^2}}+2 \\  \\ \Rightarrow x=De^{\frac{1}{5t^2}}+1
3 0
3 years ago
What is the exact circumference of the circle?
tensa zangetsu [6.8K]
Your answer would be B.)
3 0
3 years ago
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Help!! 50 points and brainliest!
Viktor [21]

Answer:

Second choice:

x=2t

y=4t^2+4t-3

Fifth choice:

x=t+1

y=t^2+4t

Step-by-step explanation:

Let's look at choice 1.

x=t+1

y=t^2+2t

I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

y=(x-1)^2+2(x-1)

Expand using the distributive property and the identity (u+v)^2=u^2+2uv+v^2:

y=(x^2-2x+1)+(2x-2)

y=x^2+(-2x+2x)+(1-2)

y=x^2+0+-1

y=x^2

So this not the desired result.

Let's look at choice 2.

x=2t

y=4t^2+4t-3

Solve the first equation for t by dividing both sides by 2:

t=\frac{x}{2}.

Let's plug this into equation 2:

y=4(\frac{x}{2})^2+4(\frac{x}{2})-3

y=4(\frac{x^2}{4})+2x-3

y=x^2+2x-3

This is the desired result.

Choice 3:

x=t-3

y=t^2+2t

Solve the first equation for t by adding 3 on both sides:

x+3=t.

Plug into second equation:

y=(x+3)^2+2(x+3)

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

y=(x^2+6x+9)+(2x+6)

y=(x^2)+(6x+2x)+(9+6)

y=x^2+8x+15

Not the desired result.

Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

x=t+1

y=t^2+4t

Solve the first equation for t by subtracting 1 on both sides:

x-1=t.

Plug into equation 2:

y=(x-1)^2+4(x-1)

Distribute and use the binomial square identity used earlier:

y=(x^2-2x+1)+(4x-4)

y=(x^2)+(-2x+4x)+(1-4)

y=x^2+2x+-3

y=x^2+2x-3.

This is the desired result.

3 0
4 years ago
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Which number has the smallest value
tester [92]

Answer:

c would be the correct answer

Step-by-step explanation:

4 0
3 years ago
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