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3 years ago

How do you find the slope and y-intercept

Mathematics

1 answer:

ankoles [38]3 years ago

7 0

To find them it's y=mx+b

b=the y intercept
m=the slope

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The total weight of samsiah and Julia is twice the weight of Mary. if samsiah weighs 54 kg and Mary weighs 60 kg, find the actua

Wewaii [24]

s + j = 2m

s = 54

m = 60

54 + j = 120

j = 120 - 54

j = 66

4 0

2 years ago

1. Construct Paasche's price index number from the following table for 2017

Crank

Answer:

Step-by-step explanation:

Tess is going to purchase a new car that has a list price of $29,190. She is planning on trading in her good-condition 2006 Dodge Dakota and financing the rest of the cost over four years, paying monthly. Her finance plan has an interest rate of 10.73%, compounded monthly. Tess will also be responsible for 7.14% sales tax, a $1,235 vehicle registration fee, and a $97 documentation fee. If the dealer gives Tess 75% of the listed trade-in price on her car, once the financing is paid off, what percent of the total amount paid will the interest be? (Consider the trade-in to be a reduction in the amount paid.) ANSWER A

4 0

3 years ago

Which of the following is the best estimate of 31% of 15? 7.5 3 5 4

Alex777 [14]

Th answer is 5 because 31% of 15 is 4.65

3 0

3 years ago

Read 2 more answers

Pls awnser i will give brainliest pls pls

Airida [17]

Answer:

Step-by-step explanation:

Divide the smaller numbers in the table by the larger numbers (in this case diameter divided by circumference) and then compare that with the choices.

5 0

3 years ago

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago