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Gelneren [198K]
3 years ago
10

Solve for s.- 6 – 2s = -6​

Mathematics
1 answer:
salantis [7]3 years ago
8 0

Answer:

s=0

-6-2(0)=-6

-6-0=-6

=6=-6

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Describe the vector that defined the following translation.<br><br> HELPPPPP RNN
siniylev [52]

Answer:

t.uy

Step-by-step explanation:

3 0
2 years ago
The population, p, of a town after t years is represented using the equation p=10000(1.04)^-t Which of the following is an equiv
Lisa [10]

Answer:

p=10000(26/25)t

Step-by-step explanation:

Considering that the population, p, of a town after t years is represented using the equation p=10000(1.04)^-t

The equation is equivalent to p=10000(26/25)t, because;

26/25 = 1.04

Hence; p=10000(26/25)t =p=10000(1.04)^t

3 0
3 years ago
Read 2 more answers
HELP!! Algebra help!! Will give stars thank u so much &lt;333
Anna35 [415]

Answers:

  • Part a)  \bf{\sqrt{x^2+(x^2-3)^2}
  • Part b)  3
  • Part c)   2.24
  • Part d)  1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying d = \sqrt{x^2 + (x^2-3)^2} is the same as writing d = sqrt(x^2-(x^2-3)^2)

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

Part (b)

Plug in x = 0 and you should find the following

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

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

Part (c)

Repeat for x = 1

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\

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

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0
3 years ago
Joy is reading a 352 page novel for her summer reading project on Monday she reads 3/8 of the novel on Tuesday she reads 28 page
MAXImum [283]

Answer:

104 Pages

Step-by-step explanation:

on Monday she reads 3/8 of the novel which means

\frac{3}{8} x 352 = 132 pages

on Tuesday she reads 28 pages, doesn't require any calculations.

on Wednesday she reads 1/4 of the novel,

\frac{1}{4} x 352 = 88

Just add all of that,

132+28+88=248 Pages

Subtract the novel pages by the read pages value

352-248=104pages.

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3 years ago
How much are 10 australian dollars worth in Canadian dollars
cupoosta [38]

Answer: 8.94

Step-by-step explanation:

6 0
3 years ago
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