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Masteriza [31]
3 years ago
13

You might need:

Mathematics
1 answer:
Nastasia [14]3 years ago
5 0

Answer:

11 + x

Step-by-step explanation:

A = LW

W = \frac{A}{L}

Plug in given information:

W = \frac{121-x^{2} }{11-x}

Factor the numerator:

W = \frac{(11+x)(11-x)}{(11-x)}

Cancel (11 - x):

W = 11 + x

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These are the first six terms of a sequence with a = 2:
SpyIntel [72]

Answer:

<h2>an+1 = 2×7ⁿ</h2>

Step-by-step explanation:

98÷14

=7

686÷98

=7

4 802÷686

=7

33 614÷4 802

=7

Then the common ratio q for this sequence is 7

recursive formula : an+1 = q×an = ?

an= a1 × qⁿ⁻¹

   =2×7ⁿ⁻¹

 

an+1 = q×an

        = 7×(2×7ⁿ⁻¹)

       = 2×7ⁿ

3 0
3 years ago
Read 2 more answers
If 2n = 54/9 then n = ______
Fantom [35]

Answer:

1). 3

2). 6

3). 9

Step-by-step explanation:

ksjsjsjsjs

8 0
3 years ago
Graph then line that represents the equation y=-2/3x +1 ?
Aleksandr [31]
First, you would start on the orgin (0,0), and move up 1. You should now be at (1,0) make a point here
Next, you will move down 2. You should now be at (-1,0)
Finally, you move right 3.

You should now be at (-1,3)
Make a point

Hope this helps!
7 0
3 years ago
WILL MARK BRAINLYEST.
gtnhenbr [62]

Answer:

1) The straight line on the graph below intercepts the two coordinate axes. The point where the line crosses the x-axis is called the [x-intercept]. The [y-intercept] is the point where the line crosses the y-axis. Notice that the y-intercept occurs where x = 0, and the x-intercept occurs where y = 0.

2) There's another important value associated with graphing a line on the coordinate plane. It's called the "y intercept" and it's the y value of the point where the line intersects the y- axis. For this line, the y-intercept is "negative 1." ... This point will always have an x coordinate of zero.

Step-by-step explanation:


8 0
3 years ago
SOLVE. integration of (1-v) /(1+v^2)
Nutka1998 [239]
\displaystyle\int\frac{1-v}{1+v^2}\,\mathrm dv=\int\frac{\mathrm dv}{1+v^2}-\int\frac v{1+v^2}\,\mathrm dv

The first integral is already standard and has an antiderivative in terms of \arctan v. For the second integral, take w=1+v^2 so that \dfrac{\mathrm dw}2=v\,\mathrm dv. Then

\displaystyle\int\frac{\mathrm dv}{1+v^2}-\int\frac v{1+v^2}\,\mathrm dv=\arctan v-\frac12\int\frac{\mathrm dw}w
=\arctan v-\dfrac12\ln|w|+C
=\arctan v-\dfrac12\ln(1+v^2)+C
4 0
3 years ago
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