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3 years ago

Complete the equation of the line through (3,-8)(3,−8)left parenthesis, 3, comma, minus, 8, right parenthesis and (6,-4)(6,−4

1 answer:

4 0

$y=mx+b$
$(3;-8)\rightarrow x=3;\ y=-8;\ (6;-4)\rightarrow x=6;\ y=-4$

$substitute:\\\\ \left\{\begin{array}{ccc}-8=3m+b\\-4=6m+b&|\cdot(-1)\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}-8=3m+b\\4=-6m-b\end{array}\right}\ \ \ |add\ both\ sides\\.\ \ \ \ -4=-3m\ \ \ |:(-3)\\.\ \ \ \ m=\dfrac{4}{3}\\\\substitute\ the\ value\ of\ m\ to\ first\ equation:\\\\3\cdot\dfrac{4}{3}+b=-8\\\\4+b=-8\ \ \ |-4\\b=-12\\\\Answer:\boxed{y=\dfrac{4}{3}x-12}$

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Alex777 [14]

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7 0

3 years ago

At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce

Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$ . By the Mean Value Theorem, there is a number c such that $0 < c$ with $v'(c)=60 \:{\frac{mi}{h^2}}$ . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly $60 \:{\frac{mi}{h^2}}$ .

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $v(0 \:h) = 30 \:{\frac{mi}{h} }$ and $v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} }$ (note that 20 minutes is $20/60=1/3$ of an hour), so the average rate of change of v on the interval $[0 \:h, \frac{1}{3} \:h]$ is

$\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in $(0 \:h, \frac{1}{3} \:h)$ at which $v'(c)=60 \:{\frac{mi}{h^2}}$ .

c is a time time between 2:00 and 2:20 at which the acceleration is $60 \:{\frac{mi}{h^2}}$ .

4 0

3 years ago

Jon has $5,000 to invest in a savings account that has interest compounded annually. If he want his money to double in eight yea

Nataly_w [17]

$5000 × 1.1%=$55

$55 × 12 mos. in a year= $660 yearly on this percent

calculate
$660 × 8 years from now =$5280

so the answer would be 1.1% I assume

6 0

3 years ago

Bobs supermarket sells red apples and green apples at the ratio of 5:3.

nika2105 [10]

Answer:

10 red apples and 6 green apples

Step-by-step explanation:

first you multiply 5×2 to get 10 apples

then you multiply 3×2 to get 6 apples

5 0

3 years ago

What is the surface area for this design?

Umnica [9.8K]

Answer:

Surface Area = 157 in^2

Step-by-step explanation:

Surface area is the area of each side added together.

Front = Trapezoid = b1 + b2 * h 5 + 9 * 5 = 27.5

2 2

Back = Trapezoid = b1 + b2 * h 5 + 9 * 5 = 27.5 27.5

2 2 27.5

Right = Rectangle = l * w 6.4 * 5 = 32 32

Left = Rectangle = l * w 5 * 5 = 25 25

Bottom = Rectangle = l * w 9 * 5 = 45 45

Top = Rectangle = l * w 5 * 5 = 25 25

157 in^2

8 0

3 years ago