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mote1985 [20]
3 years ago
12

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
Angelina_Jolie [31]
Spending money, since your amount of money is decreasing, which is negative
4 0
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:

<h2><em><u>60</u></em></h2>

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

180÷3= 60 = answer as 60 tsz is $180 and add that to 2 gsz and that totals to 208 which is the total

3 0
3 years ago
Can anybody help with this one ...... ?
VLD [36.1K]

Answer:The answer is choice 3

Step-by-step explanation:

8 0
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


5 0
3 years ago
Can y’all help me with all of these
USPshnik [31]
1- it’s

2- it’s

3- let’s

4- one’s

5- he’s
5 0
3 years ago
Read 2 more answers
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