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tatiyna
3 years ago
6

Help pls I will give brainliest! frfr

Mathematics
2 answers:
QveST [7]3 years ago
7 0

Answer:

1255 that is it so bye bye

iren [92.7K]3 years ago
3 0

Answer:

6250

Step-by-step explanation:

because it's it right sorry if am wrong

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Evaluate (-2)^6 + (-3)^2
jok3333 [9.3K]

Answer:

73

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Mrs. Little would like to set up each station to be able to make 64 pancakes. How many teaspoons of salt should be measured out
dusya [7]

Answer:

48

Step-by-step explanation:

all you do is 64 times 3/4

5 0
3 years ago
Read 2 more answers
NO LINKS OR FILES!
Archy [21]

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

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

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

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 <em>t</em> 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 <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √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 <em>v(t)</em> is the derivative of <em>s(t)</em>, 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
3 years ago
Subtract 2x 2 - 6x - 4<br> from 4x 2 - 4x + 3.
mafiozo [28]
X subtracted from y means y-x
(4x^2-4x+3)-(2x^2-6x-4)
4x^2-4x+3-2x^2+6x+4
4x^2-2x^2-4x+6x+3+4=
2x^2+2x+7
6 0
3 years ago
What do all members of the family of linear functions f(x)=1m(x+3) have in common?
dsp73
All linear functions have in common...
1. Their highest exponent is 1.
2. The graphs of the equations are lines.

When finding things in common between different types of functions, you always have to look at the two sides of math; geometry and algebra. Geometry is all the graphs, and algebra is the equations.

I hope this helps!
7 0
4 years ago
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