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sasho [114]
3 years ago
9

PLEASE HELP ME Multiply the monomials: a2x5b, −0.6axb2 and 0.6a2b3

Mathematics
1 answer:
Tanzania [10]3 years ago
8 0

Answer:

-0.36 a^{5}b^{6} x^{6}

Step-by-step explanation:

Given:

a^2x^5b,,\ -0.6axb^2,\ and\ 0.6a^2b^3

Required

Multiply

The above can be written as

a^2x^5b * -0.6axb^2\ * \ 0.6a^2b^3

Split the above expression

a^2*x^5*b * -0.6*a*x*b^2\ * \ 0.6*a^2*b^3

Collect like terms

-0.6 * 0.6 * a^2*a*a^2*b *b^2 *b^3*x^5*x

Apply first law on indices

-0.36  * a^{2+1+2} *b^{1+2+3} *x^{5+1}

-0.36  * a^{5} *b^{6} *x^{6}

-0.36 a^{5}b^{6} x^{6}

Hence, a^2x^5b * -0.6axb^2\ * \ 0.6a^2b^3 is equivalent to -0.36 a^{5}b^{6} x^{6}

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Answer:

125

Step-by-step explanation:

The ratio is 5:3 so we divide the 75 by 3 to get 1 part

1 part = 75

75 x 5 = 125

125 strawberries were sold :)

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Kai went on a bike ride. Every time he stops to drink water, his bike ride takes 3/4 of a minute longer. If he stops to drink wa
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Step-by-step explanation:

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3 years ago
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Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )
DedPeter [7]
(a) First find the intersections of y=e^{2x-x^2} and y=2:

2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}

So the area of R is given by

\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx

If you're not familiar with the error function \mathrm{erf}(x), then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line y=1 with y=e^{2x-x^2}.

1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2

So the area of S is given by

\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx
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\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx
5 0
3 years ago
Find the focus for y=x^2+4x-7
padilas [110]

ANSWER

(2,-10.75)

EXPLANATION

The given function is

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We now compare this function to

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The vertex is (2,-11).

The focus is

(2,-11+ \frac{1}{4} )

(2,- \frac{43}{4} )

(2,-10.75)

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The correct answer is 4ft

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