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3 years ago

PLEASE HELP ME!!!!!! I really need help!

Mathematics

2 answers:

Alex3 years ago

8 0

Answer:

2 to 6, 3 to 9, and 4 to 12

Mashutka [201]3 years ago

7 0

Answer:

Step-by-step explanation:

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A number increased by 5 and divided by 2

Goshia [24]

Answer:

3 is the answer shdjjdksjsjsos

4 0

3 years ago

Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect

Tasya [4]

$4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1$

defines an ellipse centered at $(0,0)$ with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the $x$ -axis. So if cross sections are taken perpendicular to the $x$ -axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at $x=x_0$ will have a base of length

$\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}$
$\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}$

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a $\sqrt2:1$ ratio with either of the remaining legs. So if the hypotenuse is $\dfrac43\sqrt{9-{x_0}^2}$ , then either leg will have length $\dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}$ .

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is $\dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}$ .

Finally, $x_0$ can be chosen from any value in $-3\le x_0\le3$ . We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

$\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)$

and so the volume would be

$\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx$
$=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}$
$=16$

6 0

3 years ago

9x9+3x3= what does that equal too

svp [43]

Answer:

252

9x9=81 +3=84

84x3=252

3 0

3 years ago

Read 2 more answers

Hi, so I know the answer to this problem (now that I got it wrong) but I'm not quite sure why I was wrong, help?

IgorC [24]

Answer:

$-2x^2-x$ . Your answer was wrong because "x-squared" terms didn't cancel.

Step-by-step explanation:

To solve this problem, set up an equation. We know something is supposed to be added to the expression $2x^2+2x$ , and the result should be x. So:

$2x^2+2x+(\text{ ? })=x$

We want to solve for the question mark... the unknown thing that we're adding to the original expression, in order to get x.

It is uncommon to put question marks in equations to represent quantities. Usually we use a letter. Since x is already being used in the equation, we should pick something else ... we could use "y".

$2x^2+2x+(\text{ } y \text{ })=x$

...or just...

$2x^2+2x+y=x$

Algebra allows us to solve the equation and find out what "y" is equivalent to.

To solve, we want to get the "y" by itself. To do so, we try to eliminate the other "terms" from the left side of the equation.

Understanding "terms" & "like terms"

Terms

"Terms" in an equation are either a number multiplied to other things, or just a single number that isn't multiplied to anything else.

For example, the various terms in our equation above are

$2x^2$ , $2x$ , $y$ , $x$

You might ask why the last things, which don't have a number, are considered terms.

Remember that multiplying by 1 doesn't change anything, so we could imagine each of the last two terms as being 1 times the letter.

So, we can rewrite our equation:

$2x^2+2x+1y=1x$

Like terms

"Like terms" are terms where the "other stuff the numbers are multiplied to" is the same, so for instance, the $2x$ and the $1x$ are like terms. They are like terms because, the "other stuff" that the numbers are multiplied to are "x" for both terms. Note that $2x$ and $2x^2$ are not "like terms" because the "stuff" is different:

$x$ is different than $x^2$

"Like terms" are important because only like terms can be "combined" into a single simplified term.

Solving equations

To solve an equation, we isolate what we're solving for, y, by disconnecting the other terms from it, and simplify.

Starting with subtracting 2x from both sides of the equation:

$2x^2+2x+1y=1x\\(2x^2+2x+1y)-2x=(1x)-2x$

Subtraction is the same as "adding a negative":

$2x^2+2x+1y+(-2x)=1x+(-2x)$

Since all terms are now connected by addition, we can add in any order we want (because of the Commutative Property of Addition), and we can combine like terms.

Thinking just about the number parts, since $1+(-2)=-1$ , then $1x+(-2x)=-1x$ .

Returning to our main equation, the right side simplifies:

$2x^2+2x+1y+(-2x)=1x+(-2x)\\2x^2+2x+1y+(-2x)=-1x$

On the left side: $2x$ and $-2x$ are like terms.

Fact: $2+(-2)=0$

So, $2x+(-2x)=0x$

Since anything times zero is just zero, $0x=0$ . Furthermore, adding zero to anything doesn't change it. So when the $2x$ and $-2x$ terms on the left side of our main equation are combined, they "disappear" (While we talked through are a lot of rules/steps to justify why that works, it is common to omit those justifications, and to just combine those like terms and make them disappear.)

So, $2x^2+2x+1y+(-2x)=-1x$ simplifies to:

$2x^2+1y=-1x$

Similarly for the $2x^2$ term, we subtract from both sides:

$2x^2+1y=-1x\\(2x^2+1y)-2x^2=(-1x)-2x^2\\2x^2+1y+(-2x^2)=-1x+(-2x^2)$

Combining like terms on the left, they disappear.

$1y=-1x+(-2x^2)$

There are no like terms on the right.

Since the two terms on the right are added together, we can use the commutative property of addition to rearrange:

$1y=-2x^2+(-1x)$

Addition of a negative can turn back into subtraction, and simplify multiplication by 1.

$y=-2x^2-x$

Remembering we chose "y" as the unknown thing we wanted to know, that's why the "correct answer" is what it is.

Verifying an answer

Verifying can double check an answer, and helps explain why the answer you chose doesn't work.

To verify an answer, the original statement said add something to the expression and get a result of "x". So, let's see if the "correct answer" does:

$2x^2+2x+(\text{ } ? \text{ })\\2x^2+2x+(-2x^2-x)\\2x^2+2x+(-2x^2-1x)\\2x^2+2x+(-2x^2)+(-1x)$

Combining the "x-squared" terms, completely cancels...

$2x+(-1x)$

Combining the "x" terms, and simplifying...

$1x\\x$

So it works.

Why isn't the answer what you chose:

$2x^2+2x+(\text{ } ? \text{ })\\2x^2+2x+(-x^2-x)\\2x^2+2x+(-1x^2-1x)\\2x^2+2x+(-1x^2)+(-1x)$

Combining the x-squared terms, things don't completely cancel...

$1x^2+2x+(-1x)$

Combining the x terms...

$1x^2+1x\\x^2+x$

So adding the answer that you chose to the expression would not give a result of "x", which is why it is "wrong"

5 0

2 years ago

Find the value of r in (4, r), (r, 2) so that the slope of the line containing them is −53

Elan Coil [88]

Answer:

C) 7

===========================================

Work Shown:

Use the slope formula

m = (y2-y1)/(x2-x1)

Plug in the given slope we want m = -5/3 and also the coordinates of the points. Then isolate r

m = (y2-y1)/(x2-x1)

-5/3 = (2-r)/(r-4)

-5(r-4) = 3(2-r) .... cross multiplying

-5r+20 = 6-3r

-5r+20+5r = 6-3r+5r .... adding 5 to both sides

20 = 6+2r

20-6 = 6+2r-6 ....subtracting 6 from both sides

14 = 2r

2r = 14

2r/2 = 14/2 .... dividing both sides by 2

r = 7

The slope of the line through (4,7) and (7,2) should be -5/3, let's check that

m = (y2-y1)/(x2-x1)

m = (2-7)/(7-4)

m = -5/3

The answer is confirmed

7 0

3 years ago

Read 2 more answers