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Aloiza [94]
3 years ago
12

Eight people were asked what the balance of their checking account at the beginning of this past week was and how much it increa

sed or decreased by the end of the week.
Create a scatter plot that represents the data that is shown in the table. The x-axis represents the beginning balance in thousands of dollars and the y-axis represents the change in the checking account in hundreds of dollars.



Name
Beginning balance

(in thousands of dollars)

Change in checking account

(in hundreds of dollars)

Billy 6 6
Barbara 4 5
Eli 3 2
Ken 1 5
Lara 8 4
Lindsey 3 8
Marie 4 2
Eduardo 5 3
Mathematics
2 answers:
TEA [102]3 years ago
6 0
Billy because she has has has beginning balance
Lapatulllka [165]3 years ago
6 0

Answer:

:D

Step-by-step explanation:

:DDDD

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Select the coordinate point that is a solution of equations. 2x+y=7; 3x-4y=5
grandymaker [24]

Answer:

D (3,1)

Step-by-step explanation:

solving simultaneous equations

Arrange equations

2x+y=7

3x-4y=5

make y side cancell by addition by multiplying by 4 and 1 as shown

4{2x+y=7}

1{3x-4y=5}

8x+4y=28

3x-4y=5

11x=33

x=33/11=3

substitute for x=3 in 3x-4y=5

9-4y=5

4=4y

y=1

4 0
3 years ago
Read 2 more answers
Hi. I need help with these questions.<br> See image for question.<br> Answer 15
Sauron [17]

Step-by-step explanation:

YOU JUST FOLLOW THE STEP

4 0
3 years ago
Read 2 more answers
Someone help please
Alla [95]

Answer:  Choice A

\tan(\alpha)*\cot^2(\alpha)\\\\

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

Explanation:

Recall that \tan(x) = \frac{\sin(x)}{\cos(x)} and \cot(x) = \frac{\cos(x)}{\sin(x)}. The connection between tangent and cotangent is simply involving the reciprocal

From this, we can say,

\tan(\alpha)*\cot^2(\alpha)\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)^2\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\frac{\cos^2(\alpha)}{\sin^2(\alpha)}\\\\\\\frac{\sin(\alpha)*\cos^2(\alpha)}{\cos(\alpha)*\sin^2(\alpha)}\\\\\\\frac{\cos^2(\alpha)}{\cos(\alpha)*\sin(\alpha)}\\\\\\\frac{\cos(\alpha)}{\sin(\alpha)}\\\\

In the second to last step, a pair of sine terms cancel. In the last step, a pair of cosine terms cancel.

All of this shows why \tan(\alpha)*\cot^2(\alpha)\\\\ is identical to \frac{\cos(\alpha)}{\sin(\alpha)}\\\\

Therefore, \tan(\alpha)*\cot^2(\alpha)=\frac{\cos(\alpha)}{\sin(\alpha)}\\\\ is an identity. In mathematics, an identity is when both sides are the same thing for any allowed input in the domain.

You can visually confirm that \tan(\alpha)*\cot^2(\alpha)\\\\ is the same as \frac{\cos(\alpha)}{\sin(\alpha)}\\\\ by graphing each function (use x instead of alpha). You should note that both curves use the exact same set of points to form them. In other words, one curve is perfectly on top of the other. I recommend making the curves different colors so you can distinguish them a bit better.

6 0
2 years ago
SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!
NARA [144]

ANSWER

Step 3

EXPLANATION

The given polynomial expression is:

2 {p}^{2}  - 3p - 7 - (3 {p}^{2}  + p - 5)

Fatau correctly expanded the parenthesis in the first step.

2 {p}^{2}  - 3p - 7 - 3 {p}^{2}   -  p  +  5

Fatau also correctly grouped the like terms to obtain:

(2- 3) {p}^{2}   + (- 3 - 1)p  + (- 7 +  5)

Fatau committed a mistake at the third step.

Instead of obtaining,

- {p}^{2} - 4p   - 2

He mistakenly got:

{p}^{2}    - 2p  +2

3 0
3 years ago
Please help with math and please show your work will give brainliest
Burka [1]

Answer:

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d,

if  and  then a=bc=da+c=b+d

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

3x+y=52x−y=0–––––––––––5x=5

The y’s add to zero and we have one equation with one variable.

Let’s try another one:

{x+4y=22x+5y=−2(5.3.3)

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2, we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.

This figure shows two equations. The first is negative 2 times x plus 4y in parentheses equals negative 2 times 2. The second is 2x + 5y = negative 2. This figure shows two equations. The first is negative 2x minus 8y = negative 4. The second is 2x + 5y = -negative 2.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Add the equations yourself—the result should be −3y = −6. And that looks easy to solve, doesn’t it? Here is what it would look like.

This figure shows two equations being added together. The first is negative 2x – 8y = −4 and 2x plus 5y = negative 2. The answer is negative 3y = negative 6.

We’ll do one more:

{4x−3y=103x+5y=−7

It doesn’t appear that we can get the coefficients of one variable to be opposites by multiplying one of the equations by a constant, unless we use fractions. So instead, we’ll have to multiply both equations by a constant.

We can make the coefficients of x be opposites if we multiply the first equation by 3 and the second by −4, so we get 12x and −12x.

This figure shows two equations. The first is 3 times 4x minus 3y in parentheses equals 3 times 10. The second is negative 4 times 3x plus 5y in parentheses equals negative 4 times negative 7.

This gives us these two new equations:

{12x−9y−12x−20y=30=28

When we add these equations,

\[{12x−9y=30−12x−20y=28–––––––––––––––––−29y=58

\]

the x’s are eliminated and we just have −29y = 58.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution

Step-by-step explanation:

Here is some examples

4 0
2 years ago
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