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Pavel [41]
2 years ago
15

I’m on my post quiz please help!!

Mathematics
1 answer:
grin007 [14]2 years ago
3 0
-5 + -7/8 because the whole thing is supposed to be negative, and if you add any kind of positive it will no longer be -5 7/8
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The Sum Of Two Numbers Is 50 And The Difference Is 18. What Are The Numbers?
mart [117]
<span>The sum of two numbers is 50. The smaller number is 18 less than the larger number. What are the numbers?
-----
Equations:
x + y = 50
x = y - 18
-------------------
Add and solve for "x"::

2x = 32
x = 16
-----
Solve for "y"::
16 + y = 50
y = 34
-----------------
Hope I helped!

~Tomas</span>
3 0
3 years ago
Andy has 31 coins made up of quarters and half dollars, and their total value is $11.00. How many quarters does he have?
charle [14.2K]

Answer:

4

Step-by-step explanation:

This situation has two unknowns - the total number of half dollars and the total number of quarters. Because we have two unknowns, we will write a system of equations with two equations using the two unknowns.

  • h+q=31 is an equation representing the total number of coins
  • 0.50h+0.25q=11 is an equation representing the total value in money based on the number of coin. 0.50 and 0.25 come from the value of a half dollar and quarter individually.

We write the first equation in terms of q by subtracting it across the equal sign to get h=31-q. We now substitute this for h in the second equation.

0.50(31-q)+0.25q=11

15.5-0.50q+0.25q=11

15.5-0.25q=11

After simplifying, we subtract 15.5 across and divide by the coefficient of q.

-0.25q=-4.5

q=4

We now know of the 31 coins that 4 are quarters.

3 0
3 years ago
Describe the location of the point having the following coordinates. negative abscissa, zero ordinate between Quadrant II and Qu
Marianna [84]

Answer:

The location of the point is between Quadrant II and Quadrant III

Step-by-step explanation:

we know that

The abscissa refers to the x-axis  and ordinate refers to the y-axis

so

in this problem we have

the coordinates of the point are (-x,0)

see the attached figure to better understand the problem

The location of the point is between Quadrant II and Quadrant III

6 0
3 years ago
−3/8 ÷ 1/2= the numbers are fractions
marishachu [46]

\text {Hi! To solve this problem you must...}

\text {Do Keep Change Flip (KCF):}

\text {Keep: -3/8}

\text {Change: /  into *}

\text {Flip: 1/2 into 2/1}

\text {Your new problem should be -3/8*2/1}

\text {The Final Step is to Multiply:}

\text {Note: When Multiplying Fractions you multiply Numerator by Numerator}

\text {and Denominator by Denominator}

\text {-3/8*2/1=}

\fbox {-6/8}

\text {Best of Luck!}

\text {-LimitedX}

4 0
2 years ago
If the function f (x) has a domain of (a,b] and a range of [c,d), then what is the domain and range of g (x) = m × f (x) + n?
xz_007 [3.2K]
<h3>Answer: Choice A</h3>

Domain = (a,b]

Range = [mc + n,md + n)

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

Explanation:

The domain stays the same because we still have to go through f(x) as our first hurdle in order to get g(x).

Think of it like having 2 doors. The first door is f(x) and the second is g(x). The fact g(x) is dependent on f(x) means that whatever input restrictions are on f, also apply on g as well. So going back to the "2 doors" example, we could have a problem like trying to move a piece of furniture through them and we'd have to be concerned about the f(x) door.

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

The range will be different however. The smallest value in the range of f(x) is y = c as it is the left endpoint. So the smallest f(x) can be is c. This means the smallest g(x) can be is...

g(x) = m*f(x) + n

g(x) = m*c + n

All we're doing is replacing f with c.

So that means mc+n is the starting point of the range for g(x).

The ending point of the range is md+n for similar reasons. Instead of 'c', we're dealing with 'd' this time. The curved parenthesis says we don't actually include this value in the range. A square bracket means include that value.

6 0
2 years ago
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