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2 years ago

6

A mountain bike originally priced at $929.00 is up for sale at a discount of 12%. What is the price of the mountain bike after t

he discount?

1 answer:

Lyrx [107]2 years ago

4 0

Answer:

Step-by-step explanation:

817.52 should be correct

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I don't know how to answer this

netineya [11]

Answer:

when the input is 3 the output is 1

Step-by-step explanation:

inputs would basically be the x coordinates while outputs are the y coordinates. when you put in the x coordinate of 3, the y coordinate of 1 is what shows up as the output. (3,1) are the coordinates on the line

7 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%20%204%20%5Ctimes%20%282%20-%206%29%20%3D%20" id="TexFormula1" title=" 4 \times (2 - 6)

Brut [27]

Answer:

-16

Step-by-step explanation:

You have to start by answering the parenthesis first then continue with the question

7 0

2 years ago

The sum of 2 numbers is 38. one number is 10 more than the other.

son4ous [18]

Answer:

24, 14

Step-by-step explanation:

x + y = 38

x = 10 + y

You can substitute (10+y) into the top equation where the x is.

10 + y + y = 38

10 + 2y = 38

2y = 28

y = 14

Now plug 14 in for y in either equation to get x

x = 10 + 14

x = 24

8 0

2 years ago

GEOMETRY HELP PLEASE!! WILL MARK BRAINLEIST!<br> 5

adell [148]

AC=BD

64 units=10x+24

64-24=10x

40=10x

X=40 divided by 10= 4

hope it helps:)))

5 0

2 years ago

Read 2 more answers

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

2 years ago