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3 years ago

A line passes through (-2, -3) and (4, -6) what is the slope of the line ?

Mathematics

2 answers:

MrRissso [65]3 years ago

8 0

The slope, or gradient, of the line is 6/-3 which is -2. The gradient is -2.

balandron [24]3 years ago

3 0

The slope is the change in Y over the change in X.

Slope = (-6 - -3) / ( 4 - -2)

Slope = (-6+3) / ( 4 +2)

Slope = -3/6

Slope = -1/2

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Which example would represent a negative number? Question 5 options: spending money gaining weight the temperature rising having

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Spending money, since your amount of money is decreasing, which is negative

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3 years ago

At a certain toy store, tiny stuffed zebras cost $3 and giant stuffed zebras cost $14. The store doesn't sell any

stepan [7]

Answer:

Step-by-step explanation:

tiny stuffed zebras(tsz)= $3

giant stuffed zebras(gsz)=$14

total sold=$208

208÷3=69.3recurring so that's incorrect as you cannot sell .3 of a zebra so keep going until you find the next whole number. (trial and error)

207÷3=69 but it cant be this as it is not the whole money made

208-14=194

194÷3=64.6recurring (cannot be this)

194-14=180

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3 years ago

Can anybody help with this one ...... ?

VLD [36.1K]

Answer:The answer is choice 3

Step-by-step explanation:

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3 years ago

A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par

lyudmila [28]

Answer:

$a(\frac{1}{5e})=5e$

Step-by-step explanation:

we are given equation for position function as

$s(t)=tln(5t)$

Since, we have to find acceleration

For finding acceleration , we will find second derivative

$s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)$

$=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t$

$=1\cdot \ln \left(5t\right)+\frac{1}{t}t$

$s'(t)=\ln \left(5t\right)+1$

now, we can find derivative again

$s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)$

$=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)$

$=\frac{1}{t}+0$

$a(t)=\frac{1}{t}$

Firstly, we will set velocity =0

and then we can solve for t

$v(t)=s'(t)=\ln \left(5t\right)+1=0$

we get

$t=\frac{1}{5e}$

now, we can plug that into acceleration

and we get

$a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}$

$a(\frac{1}{5e})=5e$

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