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1 year ago

The midpoint of AB is M (4, 4). If the coordinates of A are (2, 1), what are the

Mathematics

1 answer:

Sergio [31]1 year ago

7 0

Answer:

(6,7)

Step-by-step explanation:

Let B(x,y).

By the midpoint formula,

4 is the average of 2 and x, so x = 6.
Similarly, y = 7.

So, the coordinates of B are (6,7).

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(20x^2 – 3x – 9)/ (4x – 3) Using synthetic division

Bond [772]

Answer:

2x^2 + 2x +3

I think that this is correct

4 0

2 years ago

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A gym teacher orders 42 tennis balls. Each package contains three tennis balls. Which of the following equations represents the

navik [9.2K]

42 balls / 3 balls per package Equals 14 packages

8 0

3 years ago

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

4 0

2 years ago

Determine the radius, to the nearest inch, of a large can of tuna fish that has a volume of 66 cubic

Mnenie [13.5K]

The radius of can is 2.524 inches

Solution:

Given that large can of tuna fish that has a volume of 66 cubic inches and a height of 3.3 inches

To find: Radius

Given a large can of tuna fish, we know that can is generally of cylinder shape, we can use the volume of cylinder formula,

The volume of cylinder is given as:

$\text{ volume of cylinder}= \pi r^{2} h$$

Where,

"r" is the radius of cylinder

"h" is the height of cylinder

$\pi$ is a constant equal to 3.14

Substituting the given values in above formula,

$66 = 3.14 \times r^2 \times 3.3\\\\66 = r^2 \times 10.362\\\\r^2 = \frac{66}{10.362}\\\\r^2 = 6.369\\\\r = 2.524$

Thus the radius of can is 2.524 inches

4 0

3 years ago

NEED ANSWER ASAP- FOR FINALS

irga5000 [103]

Answer:

Height: 6 feet, Length: 14 feet, Width: 35 feet.

Step-by-step explanation:

The volume of a rectangular prism is length * width * height.

We know that the height of the pool is 6 feet and that one side is 2.5 times longer than the other. It doesn't matter which side, as long it's not the height. So let's just say the width is 2.5 times bigger than the length.

That means our equation can be rewritten as V = height * length* 2.5 * length.

Plugging in what we know, we get:

$2940=(6)(l)(2.5l)\\196=l^{2} \\l=14 feet$

Now we need to find the width.

$w=2.5l\\w=2.5(14)\\w= 35feet$

6 0

2 years ago