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vova2212 [387]
3 years ago
6

Given that P = (2, 9) and Q = (4, 14), find the component form and magnitude of vector PQ.

Mathematics
2 answers:
vredina [299]3 years ago
8 0
The magnitude of the vector is
\sqrt{2 {}^{2} + 5 {}^{2} } = \sqrt{29}
the component form is
<2,5>
Airida [17]3 years ago
8 0
To find the component form, <x2-x1, y2-y1>
We have P(2,9) meaning, x1=2, y1=9
And, Q(4,14) meaning, x2=4, y2=14
Now PQ = <4-2, 14-9> = <2, 5>

Now to find the magnitude, we use the distance formula aka the Pythagorean Theorem. a^2 + b^2 = c^2 and we have, x=2, y=5
So, 2^2 + 5^2 = c^2 =29
Therefore, the magnitude is c = sqrt of 29.
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Lady bird [3.3K]

Answer:

(e) csc x − cot x − ln(1 + cos x) + C

(c) 0

Step-by-step explanation:

(e) ∫ (1 + sin x) / (1 + cos x) dx

Split the integral.

∫ 1 / (1 + cos x) dx + ∫ sin x / (1 + cos x) dx

Multiply top and bottom of first integral by the conjugate, 1 − cos x.

∫ (1 − cos x) / (1 − cos²x) dx + ∫ sin x / (1 + cos x) dx

Pythagorean identity.

∫ (1 − cos x) / (sin²x) dx + ∫ sin x / (1 + cos x) dx

Divide.

∫ (csc²x − cot x csc x) dx + ∫ sin x / (1 + cos x) dx

Integrate.

csc x − cot x − ln(1 + cos x) + C

(c) ∫₋₇⁷ erf(x) dx

= ∫₋₇⁰ erf(x) dx + ∫₀⁷ erf(x) dx

The error function is odd (erf(-x) = -erf(x)), so:

= -∫₀⁷ erf(x) dx + ∫₀⁷ erf(x) dx

= 0

7 0
3 years ago
Elementary Algebra Skill
mr_godi [17]

The answer is a = \frac{4}{-3}

Step-by-step explanation:

<em>1. Convert the mixed fraction to an improper fraction</em>

To find the numerator, multiply the denominator by the whole number and add the numerator to it.

The denominator remains the same.

So, 2\frac{2}{3} will be \frac{8}{3}

<em>2. Now the equation is,</em>

\frac{3}{2} a - \frac{4}{3} a = \frac{10}{3} + \frac{8}{3} a

<em>3. Take LCM on both sides. </em>

For the left side, multiply the first fraction by \frac{3}{3} and multiply the second fraction by \frac{2}{2}

\frac{3*3}{2*3} a - \frac{4*3}{3*3} a = \frac{10+8a}{3}

<em>4. Solve by making a the subject</em>

\frac{9a-8a}{6} = \frac{10+8a}{3}

\frac{a}{6} = \frac{10+8a}{3}

\frac{3a}{6} =10+8a

\frac{a}{2} = 10 + 8a

a = 2(10 + 8a)

a = 20 + 16a

a-16a = 20

-15a = 20

a = \frac{20}{-15}

a = \frac{4}{-3}

Therefore, the answer is a = \frac{4}{-3}

Keyword: Equations

Learn more about equations at

  • brainly.com/question/10666510
  • brainly.com/question/4460262
  • brainly.com/question/8955867

#LearnwithBrainly

4 0
3 years ago
Help me!!!!! I really need to pass this
zlopas [31]

Answer:

Its A

Step-by-step explanation:

Tutor approved. I hope this helps you

6 0
3 years ago
Ann took a taxi home from the airport. The taxi fare was $2.10 per mile, and she gave the driver a tip of $5. Ann paid a total o
adell [148]

Answer:

$2.10x + 5= $49.10

x=21

Step-by-step explanation:

the x is used for a variable that changes and in this case the mph changes which lead x being the amount of miles that they had drove which at the end ends up multi. with the $2.10 . as for the 5 it was an initial as in it doesn't change.

for the second part

you subtract 5 from each side so you have the total of 2.10x= 44.10 then u divide 2.10 from each side and that will give you x = 21

3 0
3 years ago
03. O preço a ser pago por uma corrida de táxi inclui uma parcela fixa, denominada bandeirada, e uma parcela que depende da dist
uranmaximum [27]

Answer: d) v = 3.20 + 1.50n.

Step-by-step explanation:

Let 'v' be the price to pay in a 'n' kilometer.

Given: The price to be paid for a taxi ride includes a fixed parcel, called a flag, and a parcel that depends on the distance traveled.

Flag costs = $ 3.20 and

Each kilometre run costs = $ 1.50

Then, the amount of 'v' to pay in a 'n' kilometer race = Flag costs +   (n)× (Each kilometre run costs )

⇒ v= 3.20 + 1.50n

Hence, the correct option is  d) v = 3.20 + 1.50n.

7 0
3 years ago
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