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Sladkaya [172]
3 years ago
6

Let u = <5, 6>, v = <-2, -6>. Find -2u + 5v.

Mathematics
2 answers:
Hunter-Best [27]3 years ago
7 0
\bf \begin{cases}&#10;u = \ \textless \ 5, 6\ \textgreater \ \qquad &-2u\to -2\ \textless \ 5,6\ \textgreater \ \\&#10;&\qquad \boxed{\ \textless \ -10,-12\ \textgreater \ }\\&#10;v = \ \textless \ -2, -6\ \textgreater \ \qquad &5v\to 5\ \textless \ -2,-6\ \textgreater \ \\&#10;&\qquad \boxed{\ \textless \ -10,-30\ \textgreater \ }&#10;\end{cases}

scalar multiplication, now, add them up
Jobisdone [24]3 years ago
7 0

Answer:

<-20 , -42>

Step-by-step explanation:

u = <5, 6>, v = <-2, -6>

To find -2u + 5v, we use scalar multiplication

Multiply -2 with vector u

u = <5, 6>, -2u = -2<5,6> = <-10, -12>

v = <-2, -6>, 5v=5<-2, -6> = <-10, -30>

Now we do -2u + 5v

Add both the vectors we got

<-10, -12> + <-10, -30>

<-10+-10, -12-30>

<-20 , -42>

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Idenify the numbers that are located below -3.5 on a vertical number line
belka [17]

A number line is just that – a straight, horizontal line with numbers placed at even increments along the length. The numbers that are located below -3.5 on a vertical number line are (-∞,-3.5).

<h3>What is a number line?</h3>

A number line is just that – a straight, horizontal line with numbers placed at even increments along the length. It’s not a ruler, so the space between each number doesn’t matter, but the numbers included on the line determine how it’s meant to be used.

The numbers that are located below -3.5 on a vertical number line are all the numbers that are less than -3.5. The numbers that are located below -3.5 on a vertical number line are (-∞,-3.5).

Learn more about the Number line:

brainly.com/question/557284

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6 0
2 years ago
Find the integration of (1-cos2x)/(1+cos2x)
slega [8]

Given:

The expression is:

\dfrac{1-\cos 2x}{1+\cos 2x}

To find:

The integration of the given expression.

Solution:

We need to find the integration of \dfrac{1-\cos 2x}{1+\cos 2x}.

Let us consider,

I=\int \dfrac{1-\cos 2x}{1+\cos 2x}dx

I=\int \dfrac{2\sin^2x}{2\cos^2x}dx         [\because 1+\cos 2x=2\cos^2x,1-\cos 2x=2\sin^2x]

I=\int \dfrac{\sin^2x}{\cos^2x}dx

I=\int \tan^2xdx                      \left[\because \tan \theta =\dfrac{\sin \theta}{\cos \theta}\right]

It can be written as:

I=\int (\sec^2x-1)dx             [\because 1+\tan^2 \theta =\sec^2 \theta]

I=\int \sec^2xdx-\int 1dx

I=\tan x-x+C

Therefore, the integration of \dfrac{1-\cos 2x}{1+\cos 2x} is I=\tan x-x+C.

8 0
2 years ago
On Melissa's 6th birthday, she gets a $2000 CD that earns 7% interest compounded quarterly. If the CD matures on her 13th birthd
Alexandra [31]

Answer:

A\simeq3250.83

Step-by-step explanation:

The amount formula in compound interest is:

A=P(1+\frac{r}{n} )^{nt}

where:

P = principal amount

r = annual interest

n = number of compounding periods

t = number of years

We already know that:

P = $2000

r = 7\% = \frac{7\%}{100\%}=0.07

t = 7 (number of years from 6th to 13th bday)

n = 4 (quarterly in a year)

Then,

A=2000(1+\frac{0.07}{4} )^{(4)(7)}\\\\A=2000(1+\frac{0.07}{4} )^{28}\\\\A=3250.825792\\\\A\simeq3250.83

8 0
2 years ago
It took you 120 minutes to work 15 math problems. How long did you spend on<br> each problem
Karolina [17]

Answer:

8 minutes

Step-by-step explanation:

120÷ 15= 8

yup thats pretty much it

5 0
2 years ago
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Why is the product of 5/4 and 4 larger than 4?
zubka84 [21]
A simple answer is because 4 and 5/4 is equal to 5 and 1/4. Because 5/4 is 1/4 more than 4/4, or 1. 
5 0
3 years ago
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