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IgorC [24]
2 years ago
7

CHALLENGE PROBLEM Danny's Department Store advertises savings of

Mathematics
1 answer:
Oksanka [162]2 years ago
8 0

Discount are used to reduce the costs of items

The percent paid to determine the advertisement is incorrect

<h3>How to determine the correct percentage</h3>

The initial prices are given as: $44.00 to $152.00

And the discount is: 45% to 51%

At a discount of 45%, the price of the items would be:

Price = \$44.00 * (1 - 45\%)

Price = \$24.2

Price_2 = \$152.00 * (1 - 45\%)

Price_2 = \$83.6

At a discount of 51%, the price of the items would be:

Price = \$44.00 * (1 - 51\%)

Price = \$21.56

Price_2 = \$152.00 * (1 - 51\%)

Price_2 = \$74.48

The calculates prices show that the percent paid to determine the advertisement is incorrect

Read more about discounts at:

brainly.com/question/23103804

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Consider the equations 5x+10=30 and 5(x+10)=30. Do they have the same solution? Why or why not?
hodyreva [135]
If you simplify (take out the brackets) of this equation. 5(x+10)=30 then it would be

5 times x + 5 times 10 = 30

5x+10=30

So yes they have the same solution

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3 years ago
HELP ME PLEASE!!!
ser-zykov [4K]

parallel lines, have the same exact slope, hmmm what is the slope of y = -2/3 x + 1/3 anyway? well, low and behold, the equation is already in slope-intercept form, therefore


\bf ~~y=\stackrel{slope}{-\cfrac{2}{3}}x+\cfrac{1}{3}~~ has a slope of -2/3.


so we're really looking for a line whose slope is -2/3 and runs through 9,4.


\bf (\stackrel{x_1}{9}~,~\stackrel{y_1}{4})~\hspace{10em}slope =  m\implies -\cfrac{2}{3}\\\\\\\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-4=-\cfrac{2}{3}(x-9)\\\\\\y-4=-\cfrac{2}{3}x+6\implies y=-\cfrac{2}{3}x+10

8 0
3 years ago
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Children gender. a certain couple is equally likely to have either a boy child or a girl child. if the family has four children,
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A plane flying horizontally at an altitude of 3 miles and a speed of 500 mi/h passes directly over a radar station. Find the rat
konstantin123 [22]

Answer:

The rate at which the distance from the plane to the station is increasing is 331 miles per hour.

Step-by-step explanation:

We can find the rate at which the distance from the plane to the station is increasing by imaging the formation of a right triangle with the following dimensions:

a: is one side of the triangle = altitude of the plane = 3 miles

b: is the other side of the triangle = the distance traveled by the plane when it is 4 miles away from the station and an altitude of 3 miles

h: is the hypotenuse of the triangle = distance between the plane and the station = 4 miles                    

First, we need to find b:    

a^{2} + b^{2} = h^{2}   (1)    

b = \sqrt{h^{2} - a^{2}} = \sqrt{(4 mi)^{2} - (3 mi)^{2}} = \sqrt{7} miles

Now, to find the rate we need to find the derivative of equation (1) with respect to time:

\frac{d}{dt}(a^{2}) + \frac{d}{dt}(b^{2}) = \frac{d}{dt}(h^{2})

2a\frac{da}{dt} + 2b\frac{db}{dt} = 2h\frac{dh}{dt}

Since "da/dt" is constant (the altitude of the plane does not change with time), we have:  

0 + 2b\frac{db}{dt} = 2h\frac{dh}{dt}

And knowing that the plane is moving at a speed of 500 mi/h (db/dt):

\sqrt{7} mi*500 mi/h = 4 mi*\frac{dh}{dt}

\frac{dh}{dt} = \frac{\sqrt{7} mi*500 mi/h}{4 mi} = 331 mi/h  

Therefore, the rate at which the distance from the plane to the station is increasing is 331 miles per hour.

I hope it helps you!

4 0
3 years ago
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