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3 years ago

Determine the standard form of an equation that passes through (2,2) and (-6,4).

1 answer:

7 0

Slope = (4 - 2)/(-6 - 2)= 2/-8 = -1/4

y = mx + b
2 = -1/4(2) + b
2 = -1/2 + b
b = 5/2

equation
y = -1/4x + 5/2
4y = -x + 10
x + 4y = 10

answer
<span>A) x+4y=10</span>

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A number split evenly into 6 groups. in expression symbol.

deff fn [24]

Answered already okay. GFY.

7 0

3 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

2 years ago

samantha threw an apple out of a window from the tenth floor of her appartment building. the equation y = -16t^2+120 can be used

ella [17]

Answer:
time taken = 2.74 seconds

Explanation:
We are given that:
y = -16t² + 120 models the situation where:
y is the floor number
t is the time taken for the apple to hit the ground

We are also given that the floor number (y) is 10. Therefore, all we have to do is substitute with y in the above equation and solve for the time t as follows:
y = -16t² + 120
10 = -16t² + 120
120 - 10 = 16t²
110 = 16t²
t² = 110/16
t² = 6.875
either t = 2.74 seconds ..........> accepted
or t = -2.74 seconds .........> rejected as time cannot be in negative

Hope this helps :)

5 0

2 years ago

If the length of the hypotenuse of a right triangle is 39 millimeters and the length of one leg is 15 millimeters what is the le

Genrish500 [490]

Answer:

36 millimeters

Step-by-step explanation:

From Pythagoras theorem, the square of the hypotenuse is equal to the sum of the square of the two other legs

In mathematical terms;

a^2 = b^2 + c^2

Let a represent the hypotenuse = 39 mm and the length of one leg, say c, is 15 mm

Slotting in the values of a and b

39^2 = b^2 + 15^2

1521 = b^2 + 225

collect like terms

1521 - 225 = b^2

1296 = b^2

Take the square root of both sides

36 = b

Therefore b = 36 mm

6 0

3 years ago

Can someone please help me with that ? So confused

nasty-shy [4]

That's what it will look like

7 0

3 years ago