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pogonyaev
3 years ago
9

Ollie and 5 of his friends are going bowling. The cost for each of them is $12.30. How could you mentally go about finding the t

otal cost for the group?
Mathematics
2 answers:
Nikolay [14]3 years ago
8 0

Answer:

I don't know if this is what you're asking for exactly but you could do it by splitting up problem 12.30x5.

Step-by-step the :

Therefore,

0.30x5

= 1.5

AND

12x5

= 60

So the answer would be 1.5+60 which would be 61.50 dollars.

nika2105 [10]3 years ago
4 0

Answer:

The answer is 73.8

Step-by-step explanation:

Just multiply 12.30 by 6 5 friends and olly

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Answer:

300+90+9

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Step-by-step explanation:

399-45=354

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42 degrees

Step-by-step explanation:

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Andrews [41]
If we observe the graph it crosses x-axis at the following points:
1) x = - 2
2) x = - 5

This means, x+ 2 and x + 5 are the factors of the polynomial. A polynomial can be expressed as the product of its factors. So we can express the given polynomial as (x + 2)(x + 5)

Thus the answer to above question is option C
4 0
3 years ago
9-6=3, 6+3=9, 3+6=9 what is the missing fact family?
Taya2010 [7]

Answer:

<em>9-3=6</em>

Step-by-step explanation:

An addition/subtraction fact family has 2 addition equations and 2 subtraction ones.

The question already states both addition ones, but only one subtraction

The subtraction one stated is 9-6=3

The opposite is<em> 9-3=6</em>

<u>Hope this helps :-)</u>

7 0
3 years ago
Find the general solution to 1/x dy/dx - 2y/x^2 = x cos x, y(pi) = pi^2
Finger [1]

Answer:

\frac{y}{x^2}=\sin x+\pi

Step-by-step explanation:

Consider linear differential equation \frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)

It's solution is of form y\,I.F=\int I.F\,q(x)\,dx where I.F is integrating factor given by I.F=e^{\int p(x)\,dx}.

Given: \frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x

We can write this equation as \frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x

On comparing this equation with \frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x), we get p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x

I.F = e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}      { formula used: \ln a^b=b\ln a }

we get solution as follows:

\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C

{ formula used: \int \cos x\,dx=\sin x }

Applying condition:y(\pi)=\pi^2

\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C

So, we get solution as :

\frac{y}{x^2}=\sin x+\pi

4 0
3 years ago
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