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kramer
3 years ago
13

4. A value of 500 increases by 12% What is the new value?"​

Mathematics
1 answer:
Nonamiya [84]3 years ago
3 0

Answer:

increase percent=12%

hom much increase

500 of 12%

500×12/100=60

new value=500+60

new value=560

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You have quarters and dimes that total $2.80. your friends says its possible that the number of quarters is 8 more than the numb
Sidana [21]

\text{Let the number of dimes be x, then as per the friend the number of }\\
\text{quarters should be 8 more than dimes, so number of quarters}=x+8\\
\\
\text{we know that the value of 1 dime is }\$ 0.10 \text{ value of one quarter is }\$0.25\\
\\
\text{And since the total value of quarters and dimes is }\$2.80, \text{ so we have}\\
\\
0.10(x)+0.25(x+8)=2.80\\
\\
\Rightarrow 0.10x+0.25x+2=2.80\\
\\
\Rightarrow 0.35x=2.80-2\\
\\
\Rightarrow 0.35x=0.80

\Rightarrow x=\frac{0.80}{0.35}=2.29\\
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\text{so if we take 2 dimes then we'll have to take 10 quarters according to friend.}\\
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\text{So the statement of friend is } WRONG

6 0
3 years ago
the thrill amusement park charges an entry fee of 40$ and an additional 5$ pre ride, x. The splash water park an entry fee of 60
MrRissso [65]

45/X + 63/X= 108/X = Y, X= 108 divided by 2 = y, y=54


8 0
3 years ago
1) A line has a slope of -2 and passes through the points A(x,5), B(4,-5) and C(6,y). Find the coordinates of A and C.
siniylev [52]
1) Coordinates of A: (-1,5)  Coordinates of C: (6,-9)

2) Coordinates of A: (-1,-14) Coordinates of M: (6,7)


3 0
3 years ago
Write the equation of the line that passes through (0,3) and (-4,-1).
BlackZzzverrR [31]

Answer:

number 1

Step-by-step explanation:

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3 0
3 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
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