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3 years ago

Find the slope of the line

Mathematics

2 answers:

anyanavicka [17]3 years ago

8 0

We can use the points (-2, -2) and (2, -5) to solve.

Formula: y2-y1/x2-x1

= -5-(-2)/2-(-2)

= -3/4

Best of Luck!

egoroff_w [7]3 years ago

5 0

answer:

slope: - $\frac{3}{4}$

step-by-step explanation:

for this problem we are asked to find the slope of the line

our equation for slope is: $\frac{rise}{run}$

so using our graph we need to find both the rise and run to get from one point to another

going from the point (-2, -2) to (2, 5)

rise: -3 (it is negative because we are moving down)

run: 4

so now plug this into our slope equation: $\frac{rise}{run}$

making our equation - $\frac{3}{4}$

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bogdanovich [222]

Problem 1)

AC is only perpendicular to EF if angle ADE is 90 degrees

(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88

Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle

Triangle AED is acute (all 3 angles are less than 90 degrees)

So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF

-------------------------------------------------------------

Problem 2)

a) The center is (2,-3)

The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2

---------------------

b) The radius is 3 and the diameter is 6

From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2

where
h = 2
k = -3
r = 3

so, radius = r = 3
diameter = d = 2*r = 2*3 = 6

---------------------

c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.

Some points on the circle are

A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)

Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.

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3 years ago

At a local grocery, 26-ounces of a sport drink costs $1.30. What is the price per ounce?

anygoal [31]

Your answer would be 0.05 per ounce

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2 years ago

Alejandra gets paid 9 dollars per hour she works. If h is the total number of hours Alejandra works, which expression could be u

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3 years ago

A cell phone tower casts an 80-foot shadow. At the same time, a 6-foot fence post casts a shadow of 4 feet. What is the height o

AnnyKZ [126]

The height of the telephone tower is 120 feet.

The first step to solve this problem is to figure out what fraction of the height of the actual object the shadow is. We can find this by dividing the length of the shadow by the actual height.

4/6 = 2/3, meaning that the shadow is 2/3 of the actual height

Since we know that the shadow is 2/3 of the height, we can multiply by the reciprocal, 3/2, to find the actual height if we know the length of the shadow.

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3 years ago

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

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3 years ago