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trapecia [35]
3 years ago
15

How much greater is 300 than3?

Mathematics
2 answers:
egoroff_w [7]3 years ago
7 0
<span> 300 is 297 greater than 3</span>
GaryK [48]3 years ago
3 0
297 300 -3= A     .......
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What translates the following verbal phrase correctly to algebra:<br><br> y less than 21?
Lelu [443]
Do you mean something like y<21?
4 0
3 years ago
What is a dividen in a qoutient?
konstantin123 [22]

Answer:

The number which we divide is called the dividend. The number by which we divide is called the divisor. The result obtained is called the quotient.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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
\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve y=e^{2x-x^2} and the line y=1, or e^{2x-x^2}-1. The area of any such circle is \pi times the square of its radius. Since the curve intersects the axis of revolution at x=0 and x=2, the volume would be given by

\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx
5 0
3 years ago
Please help me!! Will get brainliest!
makvit [3.9K]
Completing the square follows the principle of taking a x^{2} +bx+c and converting it into (x+ \frac{b}{2})^2+d where d is the 'correctional number' as I like to call it - i.e. the number that converts the expanded bracket into the +c, since the expanded bracket will give us a x^{2} +b.

In this case, 2/2=1 so we have the first part: (w+1)^2.
Expanding this gives us w^2+2w+1. We need c to be 9, so we can just add 8.
Putting this together:

w^2+2w+9=(w+1)^2+8

Now we can solve it more easily.
Rearranging:
(w+1)^2=-8 \\ w+1= +-\sqrt{-8} = +-\sqrt{8}i=+-2 \sqrt{2}i   \\ \boxed{w=-1+ 2\sqrt{2}i\ or\ -1-2 \sqrt{2}i  }
5 0
3 years ago
Line n intersects lines 1 and m, forming the angles shown in the diagram below.
bija089 [108]

Answer:

B. 4.5

Step-by-step explanation:

(6x + 42)° and (18x - 12)° are alternate interior angles. If line l is parallel to line m, therefore,

6x + 42 = 18x - 12

Solve for the value of x. Combine like terms.

6x - 18x = -42 - 12

-12x = -54

Divide both sides by -12

-12x/-12 = -54/-12

x = 4.5

Therefore, x = 4.5 would prove lines l and m are parallel to each other.

7 0
3 years ago
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