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2 years ago

I'm sorry but I do not get the question here. I need help.

Mathematics

2 answers:

Roman55 [17]2 years ago

6 0

Answer:

wider and open downward

Nitella [24]2 years ago

5 0

Answer:

narrower and open upward

Step-by-step explanation:

multiplying a by 2 is stretching the graph vertically by a factor of two. essentially this pulls the graph upward, making the graph narrower. Imagine pulling on a rubber band. As you stretch it upwards, it gets taller, but it also gets narrower. This is also true of the parabola. You also know the parabola opens upwards because the equation is positive, and all positive parabola equations open upwards.

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The volume of the cylinder above is 2,314.25, and the radius of the base is 8.9 Find the volume of the cone.

olga55 [171]

Answer: $771.41\ units^3$

Step-by-step explanation:

The missing figure is attached.

The volume of an oblique cylinders and the volume of a right cylinder can be found with this formula:

$V_{cy}=\pi r^2h$

Where "r" is the radius and "h" is the height.

The volume of an oblique cone and the volume of a right cone can be found with this formula:

$V_{co}=\frac{1}{3}\pi r^2h$

Where "r" is the radius and "h" is the height.

According to the information given in the exercise, you know that the volume of the cylinder and also the radius of the cylinder and the cone ,are the following:

$V_{cy}=2,314.25\ units^3\\\\r=8.9\ units$

Therefore, in order to find the volume of the cone, you only need to multiply the volume of the cylinder by $\frac{1}{3}$ .

Then, you get:

$V_{co}=\frac{1}{3}(2,314.25\ units^3)\\\\V_{co}=771.41\ units^3$

8 0

2 years ago

Read 2 more answers

Andrea bought tacos from a food truck and left a 2 5 % 25%25, percent tip of $ 2 . 0 0 $2.00dollar sign, 2, point, 00. What was

Bingel [31]

Let us say the original price of taco before tip was x dollars.

Percentage of tip given =25 %

Amount of tip given = $2

Now we can say that 25 % of total amount = 2 dollars.

Making an equation,

$0.25x =2$

Now we have to solve for x, so dividing both sides by 0.25,

x= 8

Answer : The original price of tacos before tip was $8.

5 0

2 years ago

Read 2 more answers

Is 4 a solution of2=8

JulsSmile [24]

Answer:

Two doesn’t equal 8. Theoretically, eight is the solution...

Step-by-step explanation:

5 0

2 years ago

When both parties discuss the terms of a contract they are

marysya [2.9K]

B - Negotiating.

This is because they are communicating, they haven't yet agreed to anything or committed to the contract as neither parties have signed, thus leading to both C and D to be incorrect.

A, however, is not exactly a term used for this situation - contracting is more of a term used to say someone contracted a disease or the flu or etc.

Therefore, B would be correct as they are indeed negotiating the terms of the contract.

Hope this helps!

7 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

2 years ago