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marta [7]
2 years ago
15

Exponential functions algebra 1

Mathematics
1 answer:
meriva2 years ago
5 0
Your question is very unclear
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determine the householder transformation that annihilates all but the first entry of the vector [1 1 1 1 ]t. specifically, if (
Stolb23 [73]

Answer:

v=a+ 2e1= [3 1 1 1] x^{T}.

Step-by-step explanation:

Let a = ^{}\left[\begin{array}{ccc}\\1&1&1\\\end{array}\right] ^{T}and we know that\alpha=±║α║₂= 2.

As we have gone over in class, the formula for v is v= a-\alphae1

with the sign of chosen to avoid subtraction. This gives v=a+ 2e1= [3 1 1 1] x^{T}.

Disclaimer ;The question is incomplete

Question:

Determine the Householder transformation that annihilates all but the first entry of the vector[1 1 1 1]>Specifically, if

I-2vv^{T}/v^{T}v\left[\begin{array}{ccc}1\\1\\1\\1\end{array}\right]==\left[\begin{array}{ccc}\alpha \\0\\0\\0\end{array}\right]

what are the values of α and v?

#SPJ4

learn more about it on

brainly.com/question/28180105

8 0
1 year ago
Find angle D if angle B = 50
Mariana [72]
<h3>Answer:  80 degrees</h3>

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

Explanation:

I'm assuming that segments AD and CD are tangents to the circle.

We'll need to add a point E at the center of the circle. Inscribed angle ABC subtends the minor arc AC, and this minor arc has the central angle AEC.

By the inscribed angle theorem, inscribed angle ABC = 50 doubles to 2*50 = 100 which is the measure of arc AC and also central angle AEC.

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

Focus on quadrilateral DAEC. In other words, ignore point B and any segments connected to this point.

Since AD and CD are tangents, this makes the radii EA and EC to be perpendicular to the tangent segments. So angles A and C are 90 degrees each for quadrilateral DAEC.

We just found angle AEC = 100 at the conclusion of the last section. So this is angle E of quadrilateral DAEC.

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

Here's what we have so far for quadrilateral DAEC

  • angle A = 90
  • angle E = 100
  • angle C = 90
  • angle D = unknown

Now we'll use the idea that all four angles of any quadrilateral always add to 360 degrees

A+E+C+D = 360

90+100+90+D = 360

D+280 = 360

D = 360-280

D = 80

Or a shortcut you can take is to realize that angles E and D are supplementary

E+D = 180

100+D = 180

D = 180-100

D = 80

This only works if AD and CD are tangents.

Side note: you can use the hypotenuse leg (HL) theorem to prove that triangle EAD is congruent to triangle ECD; consequently it means that AD = CD.

5 0
3 years ago
For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[
Ivenika [448]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

Solve point 1 that is \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\:

when,

k= 1 \to  s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\

                  = \frac{1}{2} - \frac{1}{3}\\\\

k= 2 \to  s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\

                  = \frac{1}{3} - \frac{1}{4}\\\\

k= 3 \to  s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\

                  = \frac{1}{4} - \frac{1}{5}\\\\

k= n^  \to  s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\

Calculate the sum (S=s_1+s_2+s_3+......+s_n)

S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\

   =\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\

When s_n \ \ dt_{n \to 0}

=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\

\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}

In point 2: \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

when,

k= 1 \to  s_1 = \frac{1}{(1+6)(1+7)}\\\\

                  = \frac{1}{7 \times 8}\\\\= \frac{1}{56}

k= 2 \to  s_1 = \frac{1}{(2+6)(2+7)}\\\\

                  = \frac{1}{8 \times 9}\\\\= \frac{1}{72}

k= 3 \to  s_1 = \frac{1}{(3+6)(3+7)}\\\\

                  = \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\

k= n^  \to  s_n = \frac{1}{(n+6)(n+7)}\\\\

calculate the sum:S= s_1+s_2+s_3+s_n\\

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\

when s_n \ \ dt_{n \to 0}

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066

\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}

8 0
2 years ago
When Robert got a puppy from the shelter, it weigh 11 pounds. The puppy gained 40% of its original weight in the first month tha
Iteru [2.4K]
The puppy weighs 15.4 pounds, 40% of 11 is 4.4 so add that to the wight and its 15.4
3 0
2 years ago
* ANSWER PLS * Convert 55° to radians.
algol13

Answer:

11*pi/36

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
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