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3 years ago

A line is drawn so that it passes through the points (-3,0) and (1.4). What is the slope of the line?

Mathematics

1 answer:

kotegsom [21]3 years ago

3 0

Answer:

0000000000000000

Step-by-step explanation:

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Write the equation of the circle graphed below

Papessa [141]

Answer:

(x +2)^2 + (y +2)^2 = .5625

Step-by-step explanation:

(x - h)^2 + (y - k)^2 = r^2

(x + 2)^2 + (y + 2)^2 = .75^2

5 0

3 years ago

14. Molly puts two shapes together to make a new shape. The new shape has 6 sides and 6 vertices. Which two shapes did she use?

olganol [36]

She used two trapezoids.

She put one trapezoid on top and one on the bottom.

The shape she ends up making is a hexagon.

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3 years ago

2.) 4x + 8.2 ≤ 32.2 * A. x ≤ 60 B. x ≥ 6 C. x ≥ 60 D. x ≤ 6

Inga [223]

Answer:

Step-by-step explanation:

Given

4x + 8.2 ≤ 32.2 ( subtract 8.2 from both sides )

4x ≤ 24 ( divide both sides by 4 )

x ≤ 6 → D

6 0

3 years ago

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Which of the following do you need to make a box & whisker plot?

nikitadnepr [17]

Answer:

Median, box and whisker plots are all about the median

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula

Ilia_Sergeevich [38]

Answer: A. $A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}$

B. A'(5) = 1.76 cm/s

Step-by-step explanation: Rate of change measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

$A=\pi.r^{2}$

So to find the rate of the area:

$\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}$

$\frac{dA}{dr} =2.\pi.r$

Using $r(t)=3-\frac{363}{(t+11)^{2}}$

$\frac{dr}{dt}=\frac{726}{(t+11)^{3}}$

Then

$\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]$

$\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}$

Multipying and simplifying:

$\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$

The rate at which the area is increasing is given by expression $A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$ .

B. At t = 5, rate is:

$A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}$

$A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}$

$A'(5)=\frac{2408693760\pi}{4294967296}$

$A'(5)=1.76268$

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

5 0

3 years ago