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Lady bird [3.3K]
3 years ago
8

What is 7/8+14/16+4/5

Mathematics
2 answers:
Sati [7]3 years ago
6 0
The answer is 2.55 in total
Jlenok [28]3 years ago
3 0

Answer:

The answer is 2.55

Step-by-step explanation:


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Answer:

121

Step-by-step explanation:

trust me i got you wuth that 2016 throwback

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Find the absolute maximum and minimum values of the following function on the given interval. If there are multiple points in a
Bond [772]

f(x)=3\sin x+3\cos x\implies f'(x)=3\cos x-3\sin x

f has critical points where f'=0:

3\cos x-3\sin x=0\implies\cos x=\sin x\implies\tan x=1\implies x=\pm\dfrac\pi4+2n\pi

where n is any integer. We get solutions in the interval \left[0,\frac\pi3\right] for n=0, for which x=\frac\pi4.

At this critical point, we have f\left(\frac\pi4\right)=3\sqrt2\approx4.243.

At the endpoints of the given interval, we have f(0)=3 and f\left(\frac\pi3\right)=\frac{3+3\sqrt3}2\approx4.098.

So we have the extreme values

\max\limits_{x\in\left[0,\frac\pi3\right]}f=3\sqrt2

\min\limits_{x\in\left[0,\frac\pi3\right]}f=3

8 0
3 years ago
The points (3, 9) and (–3, –9) are plotted on the coordinate plane using the equation y = a • x. What is the value of a?
iren [92.7K]

Answer:

multiply

Step-by-step explanation:

5 0
3 years ago
Here’s a graph of a linear function. Write the equation that describes that function
Aleksandr [31]
It would be y= 2/4 x + 2
3 0
3 years ago
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In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En
mina [271]

Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

\frac{dy}{dx} = 3t^{2} - 2t - 27

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

3t^{2} - 2t - 27 = 0

So

a = 3, b = -2, c = -27

Then

\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328

So

t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35

t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55

y = t^{2} - 4 = (3.35)^2 - 4 = 7.22

The first point is (27.55, 7.22)

t = -2.685

x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3

y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21

The second point is (-11.3, 3.21).

8 0
3 years ago
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