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4 years ago

Help I’ll mark you brainliest plz help

Mathematics

1 answer:

Katarina [22]4 years ago

3 0

Answer:52°

Step-by-step explanation:

x+x+14=90

Collect like terms

x+x=90-14

2x=76

Divide both sides by 2

2x/2=76/2

x=38

Angle (x+14)°=(38+14)°=52° which is vertically opposite to angle PKQ.

Vertically opposite angles are equal .

Therefore angle PKQ=52

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Find a unit vector that has the same direction as the given vector. 8i ? j + 4k

e-lub [12.9K]

Answer with Step-by-step explanation:

a.Let given vector

a=8i-j+4k

$\mid a\mid=\sqrt{(8)^2+(-1)^2+(4)^2}=9$

By using the formula

Magnitude of a vector= $\sqrt{x^2+y^2+z^2}$

Where x=Coefficient of i

y=Coefficient of j

z=Coefficient of k

Unit vector= $\hat{a}=\frac{a}{\mid a\mid}$

By using the formula

The unit vector= $\hat{a}=\frac{8i-j+4k}{9}=\frac{8}{9}i-\frac{1}{9}j+\frac{4}{9}k$

b.Let vector b=-2i+4j+2k

$\mid b\mid=\sqrt{(-2)^2+4^2+2^2}=2\sqrt 6$

$\hat{b}=\frac{-2i+4j+2k}{2\sqrt 6}$

Length of vector=6

Therefore, the vector in the direction of <-2,4,2> with length 6 is given by = $6\hat{b}=6\times \frac{-2i+4j+2k}{2\sqrt 6}$

The vector in the direction of <-2,4,2> with length 6 is given by = $-\sqrt 6(-i+2j+k)$

c. $\theta=\frac{\pi}{3}$

$\mid v\mid=8$

Let v= $v_x i+v_y j$

$v_x=\mid v\mid cos\theta$

$v_x=8cos\frac{\pi}{3}=8\times \frac{1}{2}=4$

$cos\frac{\pi}{3}=\frac{1}{2}$

$v_y=\mid v\mid sin\theta$

$v_y=8\times sin\frac{\pi}{3}=8\times \frac{\sqrt 3}{2}=4\sqrt 3$

$sin\frac{\pi}{3}=\frac{\sqrt 3}{2}$

Therefore, the vector v in component form= $4i+4\sqrt 3j$

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LiRa [457]

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4 years ago

2x + y = 3

Arlecino [84]

It’s d
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3 years ago

Round 145 to the nearest hundreds place.<br> 100<br> 140<br> 150<br> 200

wariber [46]

Answer:

150

Step-by-step explanation:

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3 years ago

Read 2 more answers

Consider the line =−2x7y−1. Find the equation of the line that is parallel to this line and passes through the point −−4, 2.

skelet666 [1.2K]

Answer:

We have: 2x+7y=−1

Let’s express this equation in terms of y:

⇒y=−27x−17

⇒ Gradient =−27

Parallel lines always have the same gradient:

⇒ Gradient of parallel lines =−27

Perpendicular lines have the negative reciprocal of the gradient:

⇒ Gradient of perpendicular lines =72

Step-by-step explanation:hope it helps

8 0

1 year ago