Which shows an angle that is approximately 80°?

Written as a simplified polynomial in standard form, what is the result when

arlik [135]

8x3 + 10x2 – 8x + 1

: Hope this helps :)

5 0

3 years ago

Please help I will mark brainlist if correct

love history [14]

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Equation
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$y - 9 = - \dfrac{1}{2} (x + 2)$

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Rearrange the equation
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$y - 9 = - \dfrac{1}{2} (x + 2) \\ \\
 y - 9 = - \dfrac{1}{2}x - 1 \\ \\
 y - 9 + 9 = - \dfrac{1}{2}x - 1 + 9 \ \Longleftarrow \text{add 9 to both side} \\ \\
 y = -\dfrac{1}{2}x + 8$

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slope of the perpendicular line
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slope = 2

(*the slope of the perpendicular line is the negative reciprocal of the slope of line)

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Substitute slope into equation y = mx + c
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y = 2x + c

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Find y-intercept
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y = 2x + c
At point (0, 4)
4 = 2(0) + c
c = 4

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Substitute y-intercept into the equation y = 2x + c
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y = 2x + 4

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Answer: y = 2x + 4
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3 0

3 years ago

SOMEONE HELP ME FIND THE Y PLEASE ASAP

Evgesh-ka [11]

Answer:

(Down Below)

Step-by-step explanation:

Plug in the value of x in the left column for all the values in the equation. I will do a couple for you.

Since -4 is the value of x, substitute it into the equation for x.

$y= -4-5=-9$

-1

$y=-1-5=-6$

0

$y=0-5=-5$

You can do the rest. You got this! Comment if you need more help!

3 0

2 years ago

Read 2 more answers

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Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

(c) The particle is at rest when its velocity is zero:

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago

Solve this equation.

spin [16.1K]

Answer:

D: 6 5/6

Step-by-step explanation:

Solve for a:

a + 7/6 = 8

Put each term in a + 7/6 over the common denominator 6: a + 7/6 = (6 a)/6 + 7/6:

(6 a)/6 + 7/6 = 8

(6 a)/6 + 7/6 = (6 a + 7)/6:

(6 a + 7)/6 = 8

Multiply both sides of (6 a + 7)/6 = 8 by 6:

(6 (6 a + 7))/6 = 6×8

(6 (6 a + 7))/6 = 6/6×(6 a + 7) = 6 a + 7:

6 a + 7 = 6×8

6×8 = 48:

6 a + 7 = 48

Subtract 7 from both sides:

6 a + (7 - 7) = 48 - 7

7 - 7 = 0:

6 a = 48 - 7

48 - 7 = 41:

6 a = 41

Divide both sides of 6 a = 41 by 6:

(6 a)/6 = 41/6

6/6 = 1:

Answer: a = 41/6 or a= 6 5/6

7 0

3 years ago