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nadezda [96]
3 years ago
8

50*50+60+90+40-60-43-89-30-34+12​

Mathematics
2 answers:
tensa zangetsu [6.8K]3 years ago
5 0
The answer is 2446.
White raven [17]3 years ago
3 0
50x50=2500
60+90+40=190
-60-43-89-30-34+12= -244
2500+190-244=2446
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Find the length of the missing
Ivahew [28]

Answer:

12

Step-by-step explanation:

using pythagoras theorem

here 15 is hypotenuse since it is opposite 0f 90 degree

9 and x are the other smaller sides of a triangle

according to pythagoras thorem the sum of square of two smaller sides of a triangle is equal to the square of hypotenuse. So,

9^2 + x^2 = 15^2

81 + x^2 = 225

x^2 = 225 - 81

x^2 = 144

x = \sqrt{144

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5 0
3 years ago
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Which statements can be represented by this equation? Select three options.<br> 4<br> 29.2 - 14
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6 0
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Rudik [331]

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7 0
3 years ago
The graph translated horizontally two units right, vertically four units down and reflected across x-axis from its parent functi
Keith_Richards [23]

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Right 2 units:                    f(x) = (x^2 - 2)

Down 4 units:                   f(x) = (x^2 - 2) - 4

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Answer:                            f(x) = - (x^2 -2) + 4

3 0
3 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
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