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EleoNora [17]
3 years ago
15

If a video game was 10% off and the amount of the discount was $6 what was the original price?

Mathematics
1 answer:
Alenkinab [10]3 years ago
4 0

Answer: the original price was $60! <3

Step-by-step explanation:

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I think it is 5 but I am not for sure
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Guy's y'all plzz answer this who knows about it...and also exam is near​
Yanka [14]

Answer:

Step-by-step explanation:

Okay so I will explain the first one.

2 to the power of 4 = 2x2x2x2

2 to the power of 6 = 2x2x2x2x2x2

2 to the power of 4 = 2x2x2x2

2^4 = 16

2^6 = 64

When you are multiplying powers you add.

+

When you subtract you takeaway.

-

So you would do

4+6+4

= 14

Meaning 2 to the power of 14

2x^{14}

So what is 2 to the power of 14.

You would do

2x2x2x2x2x2x2x2x2x2x2x2x2x2

The answer is 16384

You can also use a calculator

GOOD LUCK FOR YOUR TEST!

7 0
3 years ago
Marc bought a giant submarine sandwich for his party. The sandwich was cut into 6 pieces. Each piece was 1/2 foot long.
KonstantinChe [14]
It was 3 feet long...............
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3 years ago
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If you add the lengths of the focal radii of an ellipse, what other value will you produce?
ss7ja [257]

Equation of an ellipse

→having center (0,0) , vertex ((\pm a ,0) and covertex (\pm b ,0) and focus (\pm c ,0) is given by:

\frac{x^2}{a^2} + \frac{y^2}{b^2}=1

As definition of an ellipse is that locus of all the points in a plane such that it's  distance from two fixed points called focii  remains constant.

Consider two points (a,0) and (-a,0) on Horizontal axis of an ellipse:

Distance from (a,0) to (c,0) is = a-c = F_{1}

Distance from (-a,0) to (c,0) is = a + c = F_{2}

F_{1} + F_{2} = a -c + a +c

          = a + a

        = 2  a  →(Option A )


8 0
3 years ago
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Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

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

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(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

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
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