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2 years ago

Given vector u, what is the magnitude, |u|, and directional angle, θ, in standard position?

Mathematics

2 answers:

Simora [160]2 years ago

6 0

Answer:

Step-by-step explanation:

Edge 2022

Dominik [7]2 years ago

5 0

The magnitude of a vector has to do with the vector length that is used to show the initial and final position of the vector.

<h3>What is a Vector?</h3>

This refers to the quantity that has a magnitude and direction and can be used to show position.

Hence, we can see that the vector, magnitude, and the directional angle values are not given so a general overview was given.

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<img src="https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2B%20%20%5Cfrac%7Bw%7D%7B3%7D%20" id="TexFormula1" ti

ElenaW [278]

Answer: $\bold{w=\dfrac{6y-3}{2}}$

<u>Step-by-step explanation:</u>

Isolate w by performing the following steps

Multiply by 6 on both sides to clear the denominator
Subtract 3 from both sides
Divide both sides by 2

$y=\dfrac{1}{2}+\dfrac{w}{3}\\\\\\6\bigg[y=\dfrac{1}{2}+\dfrac{w}{3}\bigg]\quad \implies \quad 6y=3+2w\\\\\\6y-3=3-3+2w\quad \implies \quad 6y-3=2w\\\\\\\dfrac{6y-3}{2}=\dfrac{2w}{2}\quad \implies \quad \large\boxed{\dfrac{6y-3}{2}=w}$

5 0

3 years ago

Help with 8 9 and 10

Rainbow [258]

9. 2.86600.... that is all i know so far

6 0

3 years ago

HELP ASAP!!! PLEASE CHECK ALL OF THEM!

pochemuha

Answer:

A. (P+h)(x)

2x/x+4 (x-1) + x/x-1 (x+4)

2x^2-1/x^2-4

X^2+4/x^2-4

= 3x^2+3/x^2-4

B. (F-g)(x)

X^2-7x+6-x - 6

= x^2 -8x

C. (Fg)(x)

(X^2-7x+6)(x-6)

= x^3-13x^2+48x-36

D. (H/p)(x)

X/x-1 / 2x/x+4

X/x-1 / x+4/2x

= X^2+4x/2x^2-2x

A. (F+g)(3)

X^2+1 + x-4

3^2+1 + 3-4

10 -1

= 9

B. (f-g)(0)

X^2+1 - x-4

0+1 -0-4

1-4

= -3

C. (Fg)(-k)

(X^2+1) (x-4)

(-k^2+1) (-k-4)

K^3+4k^2-k-4

D. (F/g)(k-2)

X^2+1 /x-4

K-2^2+1 / k-2 -2

= K^2-4k+5 / k-4

Step-by-step explanation:

8 0

3 years ago

The hypotenuese of an isosceles right triangle is 13 inches. The midpoints of its sides are connected to form an inscribed trian

Georgia [21]

Answer:

The Sum of the areas of theses triangles is 169/3.

Step-by-step explanation:

Consider the provided information.

The hypotenuse of an isosceles right triangle is 13 inches.

Therefore,

$x^2+x^2=169\\2x^2=169\\x=\frac{13}{\sqrt{2} }$

Then the area of isosceles right triangle will be: $A=\frac{1}{2} x^2$

Therefore the area is: $A=\frac{169}{4}$

It is given that sum of the area of these triangles if this process is continued infinitely.

We can find the sum of the area using infinite geometric series formula.

$S=\frac{a}{1-r}$

Substitute $a=\frac{169}{4} \ and\ r=\frac{1}{4}$ in above formula.

$S=\frac{\frac{169}{4}}{1-\frac{1}{4}}$

$S=\frac{\frac{169}{4}}{\frac{3}{4}}$

$S=\frac{169}{3}$

Hence, the Sum of the areas of theses triangles is 169/3.

7 0

2 years ago

Find the future value of a 10-year investment of $3500 at a simple annual rate of 3.49%.

Setler79 [48]

Answer:

You can plug in the simple interest formula A=P(1+rt) to get the amount is equal to 3500 (the principal in this case) times 1+ 3.49/100(10). This is 3500(1.349), and this is $4721.5.

4 0

3 years ago