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Vika [28.1K]
3 years ago
6

If a bored game was originally $25 and it is on sale for $18 , what is the percent discount?

Mathematics
1 answer:
Blizzard [7]3 years ago
7 0
Hi there!

To find the percent discount, we need to divide the new price by the original price without the discount:

18 / 25 = 0.72

Now, we convert this into a percentage, so we can find the <em>percent </em>discount:

0.72 = 72%

Now, we subtract from 100%:

100% - 72% = 28%

So, the answer is 28%. 

Hope this helps!
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A kayak rental company charges $25.00 to rent a kayak and $3.50 for each half hour it is used.
pogonyaev

Answer:

Option D (r(t) = 3.50t +25 ; r(8) = 53)

Step-by-step explanation:

The fixed cost to rent the kayak $25. This is the cost which remains fixed irrespective of the usage of the kayak. The variable cost of using the kayak is the cost which depends on the usage of the kayak. It is mentioned that the kayak is used for 4 hours and the company charges $3.5 for every half hour. The cost function is given by:

r(t) = 25 + 3.5t ; there r is the total cost of using the kayak and t is the number of half-hours the kayak is used.

4 hours means that there are 8 half-hours. Therefore, t=8. Put t=8 in r(t).

r(8) = 25 + 3.5*(8) = 25 + 28 = 53.

Therefore, Option D is the correct answer!!!

3 0
3 years ago
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Guyssssssss help me 10 points!!
zlopas [31]

Answer:

19

Step-by-step explanation:

the side is same as 57

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3 years ago
PLEASE HELPP!! WILL GIVE BRAINLIEST + 25 POINTS!!
enyata [817]

Answer:

x = -9/4

Step-by-step explanation:

−2(x + 1/4) + 1 = 5 ​

Subtract 2 from both sides.

−2(x + 1/4) = 4 ​

Divide both sides by -2.

x + 1/4 = -2

Subtract 1/4 from both sides.

x = -2 1/4

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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}.

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3 years ago
How do I find cube roots
anyanavicka [17]

Answer:

Step-by-step explanation:

5-500 is sqare 5

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