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zepelin [54]
3 years ago
14

Today everything at a store is on sale The store offers a 20% discount The regular price of a tshirt is 18$ what is the discount

price
Mathematics
1 answer:
kirill [66]3 years ago
6 0
The discounted price is $14.40. Hope this helped.
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Step-by-step explanation:


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An indoor sport exhibition is coming to the arena. Your supervisor has asked you to help set up a handball pitch and seating are
Sphinxa [80]

Answer:

Perimeter: 174.8 m

Area: 1,394 sq m

Step-by-step explanation:

First, the perimeter.

Before we start, let's calculate the circumference of the half-circles at the ends of the field.

The measurement says 2,000 cm, so let's convert it to 20 m for ease.

Circumference of a circle: πd, where d = diameter.. in our case d = 20 m

Circumference of a 20m diameter circle: 20π = 62.8 m

We have 2 half circles... so the perimeter of each half-circle will be: 31.4 m

We also have 800 cm measurement for the "height" of the seating areas... let's convert that in 8 m

We also need to find out the space between the seating area...  We know the whole rectangular pitch is 40 m, then we have to subtract the width of both seating areas (20 and 15 m)... so the space between them is 5m

So, starting with the upper left corner of the rectangular pitch, and working our way clockwise, we encounter the following lengths:

P = 40 + 31.4 + 8 + 10 + 8 + 5 + 8 + 25 + 8 + 31.4 = 174.8 m

The total perimeter is then of 174.8 m

For the area, we need to calculate the area of all forms:

Large rectangular pitch:

LR = 40 x 20 = 800 sq m

The two half circles, form a circle, so A = πr², where r is the radius, which is half the diameter.

AC = π (10)² = 100 π = 314 sq m

Then the seating areas:

SA1 = 25 x 8 = 200 sq m

SA2 = 10 x 8 - 80 sq m

Then, we add up everything:

TA = LR + AC + SA1 + SA2

TA = 800 + 314 + 200 + 80 = 1,394 sq m

3 0
4 years ago
Find dy/dx if y =x^3+5x+2/x²-1
stiks02 [169]

<u>Differentiate using the Quotient Rule</u> –

\qquad\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\

According to the given question, we have –

  • f(x) = x^3+5x+2
  • g(x) = x^2-1

Let's solve it!

\qquad\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2)  \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5)  -  ( x^3+5x+2) 2x}{(x^2-1)^2 }\\

\qquad\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\

\qquad\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\

\qquad\pink{\therefore  \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}}  =  \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\

7 0
2 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B5%20%7D%20%2B%20%20%5Cfrac%7B3%7D%7B20%7D%20%20%3D%20" id="TexFormula1" ti
____ [38]
The answer is 19/20
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