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Triss [41]
3 years ago
7

A sporting goods store manager was selling a ski set for a certain price. The manager offered the markdowns​ shown, making the​

one-day sale price of the ski set ​$324. Find the original selling price of the ski set.
Mathematics
2 answers:
inn [45]3 years ago
8 0

Answer:

$520.632

Step-by-step explanation:

grigory [225]3 years ago
6 0
520 and some change
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The equation y = 10x shows how the
lubasha [3.4K]
$80.00 hope this helps!
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How do you solve b+(–2)=0
vitfil [10]
B + (-2) = 0

Here, we need to do the inverse operation, and apply it to BOTH sides of teh equation.

b -2 = 0
  +2    +2

b = 2

Now we have our answer, b = 2

Final answer: b = 2
8 0
3 years ago
Read 2 more answers
Which of the following points is a solution to y ≤ −2x + 3?
Alika [10]

Answer:

d; below; y = −2x + 2

Step-by-step explanation:

the equality  ≤  means left to right that y is equal or less than. We see that 2 is equal or less than 3.

We check the resulting equation also and see that if we +3 it = 1 and if we +2 it equals 0 so we know +2 in answer format can be less than also. But with questions we are measuring the units and +2 is less than +3.

5 0
3 years ago
CAN SOMEONE PLEASE HELP ME WORTH 20 points
Ainat [17]

Answer:

  12 units

Step-by-step explanation:

Point H is the intersection of the medians, so divides each median into parts with the ratio 2:1. Here, that is ...

  CH : HG = 2 : 1 = CH : 6

Then CH = 2×6 = 12.

_____

Alternatively, you could determine altitude CG from the Pythagorean theorem applied to triangle BCG. You would compute ...

  CG² +GB² = BC²

  CG = √(BC² -GB²) = √(19.8² -8²) = √146.7 ≈ 12.1

This is an approximation based on the fact that the value 9.9 for BF is an approximation. Based on BG and GH being integer values, the appropriate measure of BF is (√388)/2 ≈ 9.8489.

(The triangle shown cannot exist.)

5 0
3 years ago
Find the length of the height of the right trapezoid shown below, if it has the greatest possible area and its perimeter is equa
Artyom0805 [142]

Answer:

The height of the right trapezoid is   \frac{6}{5+\sqrt{3}}\ units

Step-by-step explanation:

Let

x ----> the height of the right trapezoid in units

we know that

The perimeter of the figure is equal to

P=AB+BC+CD+DH+HA

we have

P=6\ units

AB=BC=CH=HA=x ---> because is a square

substitute

6=x+x+CD+DH+x

6=3x+CD+DH -----> equation A

<em>In the right triangle CDH</em>

sin(30\°)=\frac{CH}{CD}

sin(30\°)=\frac{1}{2}

so

Remember  that CH=x

\frac{1}{2}=\frac{x}{CD}

CD=2x

tan(30\°)=\frac{CH}{DH}

tan(30\°)=\frac{\sqrt{3}}{3}

so

\frac{\sqrt{3}}{3}=\frac{x}{DH}

DH=x\sqrt{3}

substitute the values in the equation A

6=3x+CD+DH -----> equation A

CD=2x

DH=x\sqrt{3}

6=3x+2x+x\sqrt{3}

6=5x+x\sqrt{3}

6=x[5+\sqrt{3}]

x=\frac{6}{5+\sqrt{3}}\ units

5 0
3 years ago
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