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liubo4ka [24]
3 years ago
15

Choose an equation that represents the line passing through the point (2, -4) with a slope of 1/2.

Mathematics
1 answer:
olga55 [171]3 years ago
7 0
You use the formal y-y1=m(x-x1) to solve this and you plug in the #s so 2 and -4 will be x1 and y1 and 1/2 will be plunges is as m
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If set A contains seven distinct numbers and set B contains three distinct letters, how many elements are in (A U B)?​
Ira Lisetskai [31]
<h3>Answer:  10 elements</h3>

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

Explanation:

Let's use an example to see why this is

A = {1,2,3,4,5,6,7}

B = {x,y,z}

A U B = {1,2,3,4,5,6,7, x,y,z}

We simply glue the two sets together to form one single big set. Because sets A and B have no overlap, this means we dont have to worry about tossing out duplicates (since there aren't any).

There are 7 items in set A, and 3 items in set B.

Therefore, we have 7+3 = <u>10 items in set A U B</u>

The symbol U refers to the set union operator.

If you wanted to make a Venn Diagram, then you'd have two circles overlapping. In circle A will be the numbers 1 through 7. These values are not in the overlapped region. For circle B, the letters x,y,z are inside but not in the overlapped region. The overlapped region remains empty.

The set A U B talks about the collection of everything mentioned. It's the set of stuff in set A or in set B or in both sets.

-----------------

If you want a formula to write down on a notecard or reference sheet, then it would be

n(A U B) = n(A)+n(B)

where A and B are mutually exclusive, aka disjoint. This means they have nothing in the overlapped region

4 0
2 years ago
A club raises $3,207 to buy stuffed animals for a local children's hospital. Each stuffed animal costs $17. What is the maximum
Ksju [112]

Answering:

188

Explaining:

To solve this problem, we must divide the total amount of money raised by the cost of the stuffed animals. Each stuffed animal costs $17. The club raised $3,207 to buy said stuffed animals. By dividing the money earned, which is also the money the club is able to spend, by the cost of a single/one stuffed animal, we will get how many stuffed animals the club can purchase with the money they currently possess. Our equation will look like this: 3,207 ÷ 17.

After dividing 3,207 by 17, we have the number 188.64705882. This can be rounded to the nearest tenth to create the simpler yet still accurate number 188.6.

Our final step is to round 188.6 down to the whole number it already has. (That is to say, simply cut off the fraction and remove it to get our answer.) This step must be done because we are buying stuffed animals in a real-world situation. The club would not be able to purchase part of a stuffed animal for a fraction of the cost, and the cost of the stuffed animals in the problem is a fixed value. This means that the fraction is irrelevant since we cannot purchase anything with it, effectively making it totally irrelevant to the answer. After removing the fraction from 188.6, we are left with 188.

Therefore, the maximum number of stuffed animals the club can buy is <em>188 stuffed animals</em>.

4 0
3 years ago
1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a
katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

8 0
2 years ago
I you were adding a number to 189 to equal 656, what number would have to be added to the ones place of 189?
Dominik [7]

Answer:

Step-by-step explanation:

6 0
3 years ago
Help Me PLEASE!! Urgent! Marking Brainliest!!
denis-greek [22]

Answthe 2 at the bottom go to Not a polynomial

Step-by-step explanation:

5 0
3 years ago
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