PLEASE HELP ME‼️ what is the correct reason for each step and the proof?

Did I do the first question right? can anyone please help me with the 2nd question? Would really appreciate the Help

IgorC [24]

Answer:

8. 1483.33

9. 21 months

Step-by-step explanation:

8. Morgan's income after taxes is 55000/12 = 4583.33 per month. The amount available after expenses is 4583.33 -3100.00 = 1483.33 per month.

Morgan is able to put $1483.33 per month into savings.

__

9. If Morgan is able to save $1483.33 per month, it will take her ...

$30,000/$1483.33 ≈ 20.2

months to save $30,000. After 20 months, she won't have quite enough, so it will take her one more month to save the desired amount.

It will take Morgan about 21 months to save $30,000.

_

If you like, you can write an equation for "m", the number of months it will take Morgan to save 30,000:

1483.33×m = 30,000

m = 30,000/1483.33 ≈ 20.2 . . . . . . divide by the coefficient of m

7 0

3 years ago

Rebecca and dan are biking in a national park for three days they rode 5 3/4 hours the first day and 6 4/5 hours the second day

likoan [24]

Answer:

Rebecca and Dan need to ride $7\frac{9}{20}\ hrs.$ on the third day in order to achieve goal of biking.

Step-by-step explanation:

Given:

Goal of Total number of hours of biking in park =20 hours.

Number of hours rode on first day = $5\frac34 \ hrs.$

So we will convert mixed fraction into Improper fraction.

Now we can say that;

To Convert mixed fraction into Improper fraction multiply the whole number part by the fraction's denominator and then add that to the numerator,then write the result on top of the denominator.

$5\frac34 \ hrs.$ can be Rewritten as $\frac{23}{4}\ hrs$

Number of hours rode on first day = $\frac{23}{4}\ hrs$

Also Given:

Number of hours rode on second day = $6\frac45 \ hrs$

$6\frac45 \ hrs$ can be Rewritten as $\frac{34}{5}\ hrs.$

Number of hours rode on second day = $\frac{34}{5}\ hrs.$

We need to find Number of hours she need to ride on third day in order to achieve the goal.

Solution:

Now we can say that;

Number of hours she need to ride on third day can be calculated by subtracting Number of hours rode on first day and Number of hours rode on second day from the Goal of Total number of hours of biking in park.

framing in equation form we get;

Number of hours she need to ride on third day = $20-\frac{23}{4}-\frac{34}{5}$

Now we will use LCM to make the denominators common we get;

Number of hours she need to ride on third day = $\frac{20\times20}{20}-\frac{23\times5}{4\times5}-\frac{34\times4}{5\times4}= \frac{400}{20}-\frac{115}{20}-\frac{136}{20}$

Now denominators are common so we will solve the numerator we get;

Number of hours she need to ride on third day = $\frac{400-115-136}{20}=\frac{149}{20}\ hrs \ \ Or \ \ 7\frac{9}{20}\ hrs.$

Hence Rebecca and Dan need to ride $7\frac{9}{20}\ hrs.$ on the third day in order to achieve goal of biking.

8 0

3 years ago

The angle marked a = 75°. Work out the value of x. The diagram is not drawn accurately.

NikAS [45]

Answer:

x+75=180

x=180-75

x=105

Step-by-step explanation:

x and a are on a straight line

a straight line=180

so you add the x to a=180

4 0

2 years ago

a sixth grade class has 12 boys and 24 girls consider this statement: for every two boys there are 4 girls do you ag

Nitella [24]

So the ratio of boys to girls is 12:24.. now, is that an equivalent ratio as 2 boys for 4 girls? or 2:4? let's see.

$\bf \cfrac{boys}{girls}\qquad 12:24\qquad \cfrac{12}{24}\implies \boxed{\cfrac{1}{2}}
\\\\\\
\cfrac{boys}{girls}\qquad 2:4\qquad \cfrac{2}{4}\implies \boxed{\cfrac{1}{2}}$

you could start simplifying the 12/24 fraction by dividing by some common multiple, say 2, and get 6/12, then divide by 3 and get 2/4 and stop there.

then you'll notice that 12/24 is really 2/4 in disguise.

4 0

3 years ago

Justin is constructing a line through point Q that is perpendicular to line n. He has already constructed the arcs shown. He pla

krek1111 [17]

B. It must be the same as when he constructed the arc centered at point A. This problem would be a lot easier if you had actually supplied the diagram with the "arcs shown". But thankfully, with a few assumptions, the solution can be determined. Usually when constructing a perpendicular to a line through a specified point, you first use a compass centered on the point to strike a couple of arcs on the line on both sides of the point, so that you define two points that are equal distance from the desired intersection point for the perpendicular. Then you increase the radius of the compass and using that setting, construct an arc above the line passing through the area that the perpendicular will go. And you repeat that using the same compass settings on the second arc constructed. This will define a point such that you'll create two right triangles that are reflections of each other. With that in mind, let's look closely at your problem to deduce the information that's missing. "... places his compass on point B ..." Since he's not placing the compass on point Q, that would imply that the two points on the line have already been constructed and that point B is one of those 2 points. So let's look at the available choices and see what makes sense. A .It must be wider than when he constructed the arc centered at point A. Not good. Since this implies that the arc centered on point A has been constructed, then it's a safe assumption that points A and B are the two points defined by the initial pair of arcs constructed that intersect the line and are centered around point Q. If that's the case, then the arc centered around point B must match exactly the setting used for the arc centered on point A. So this is the wrong answer. B It must be the same as when he constructed the arc centered at point A. Perfect! Look at the description of creating a perpendicular at the top of this answer. This is the correct answer. C. It must be equal to BQ. Nope. If this were the case, the newly created arc would simply pass through point Q and never intersect the arc centered on point A. So it's wrong. D.It must be equal to AB. Sorta. The setting here would work IF that's also the setting used for the arc centered on A. But that's not guaranteed in the description above and as such, this is wrong.

8 0

3 years ago

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