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3 years ago

What is the horizontal change between two points on a line

Mathematics

2 answers:

Finger [1]3 years ago

5 0

Answer:

Run

Step-by-step explanation:

Helga [31]3 years ago

4 0

The horizontal change is called the run and the vertical is rise. so for example if one point to the other rises by 4 and runs by 5, the slope is 4/5. slope is
rise / run

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X:2 (105-275:11):4=28

balandron [24]

X:2+(105-275:11):4=28
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X:2+20=28
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4 0

3 years ago

Starting at 1:00 pm, the temperature changes -4 degrees Fahrenheit per hour. Write and solve an equation to find how many hours

vesna_86 [32]

Answer:

im guessing it will take 3 hours to get to -1 F

Step-by-step explanation:

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2 years ago

Find the sum. 3 6 9 ... 294

MArishka [77]

S=((294+3)*(294/3))/2=14553

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2 years ago

Assume the cost of a car is $25,000. With continuous compounding, in effect, the cost of the care will increase according to the

Masja [62]

The value of t that makes the factor e^(.032t) have the value of 2 can be found using logarithms.
2 = e^(0.032t)
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It would take 21.66 years for the cost to double.

4 0

2 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

1 year ago