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sukhopar [10]
2 years ago
14

NO LINKS..... There are 39 people on a city bus. The ratio of adults to children is 10:3. At the next stop, 3 adults get off the

bus. What is the new ratio of adults to children on the bus? A. 3:1 B. 9:2 C. 7:3 D. 12:1
Mathematics
1 answer:
Zarrin [17]2 years ago
8 0
Since the ratio is 10:3 we can find the total number of children and adults by finding how many times 13 (10+3) goes into 39. 39 / 13 = 3 so we can take the original 10:3 ratio and just scale it up to fit the whole group 10:3 * 3 -> 30:9:

From here it gets a little easier we know three adults of the 30 get off at the next stop bringing us to 27. So now our ratio is 27:9 we can simplify that down to 3:1 to get our final answer. Hope that helps!
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BabaBlast [244]

Question 3

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Question 1

if angle m is 80 degrees the m = 80 degrees

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Question 5

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Question 2

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4 0
11 months ago
At Billy's school, 80 students come to school by bicycle or by car. Together, the vehicles they arrive to school in have 270 whe
Anna35 [415]

Answer:

x = number of bicycles = 35

y = number of cars = 55

Step-by-step explanation:

Let

x = number of bicycles

y = number of cars

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2x + 4y = 270 (2)

From (1)

x = 80 - y

Substitute x = 80 - y into (2)

2x + 4y = 270 (2)

2(80 - y) + 4y = 270

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6 0
3 years ago
1 substitute p=5, q=7, r = -2
lora16 [44]

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3 years ago
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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 –

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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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