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3 years ago

Which statement is true about this equation? y=2x+4

Mathematics

2 answers:

kolezko [41]3 years ago

6 0

Step-by-step explanation:

C. is the right answer

It is a function only.

alukav5142 [94]3 years ago

4 0

Answer:

Answer and Explanation:

In solving y = 2x - 4, we find that the slope of the line is 2, the y-intercept of the line is y = -4, and the x-intercept of the line is x = 2.

I hope this helps.

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3 years ago

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A 3-gallon bottle of fabric softener costs $28.92. What is the price per quart?

bixtya [17]

Answer:

$2.41

Step-by-step explanation:

There are 4 quarts in a gallon. If there are 3 gallons that means there is 12 quarts (3x4=12) You then take $28.92 and divide it by 12, which will give you 2.40833333 and on and on, you round that to $2.41. I hope that helps!

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3 years ago

A basketball player shoots when she is 16 ft from the backboard. Given the ball has an initial velocity v0 at an angle of 30° wi

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2 years ago

A rectangular ground is 40m long 35m board 2m path is constructed inside it find its area

iren2701 [21]

Answer:

2800

Step-by-step explanation:

solution:

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A=l×b×h

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4 0

3 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

3 years ago