All categories

2 years ago

Exponential functions algebra 1

Mathematics

1 answer:

meriva2 years ago

5 0

Your question is very unclear

You might be interested in

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- $\alpha$ e1

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 $s_n \ \ dt_{n \to 0}$

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