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3 years ago

The components of a vector are Ax = –10 units and Ay = 8 units. What angle does this vector make with the positive x axis? g

Mathematics

1 answer:

Mandarinka [93]3 years ago

7 0

Answer:

This vector make an angle of 141° with the positive x axis

Step-by-step explanation:

Refer the attached figure

Components of vector :

$A_x=-10 \\A_y = 8$

We are supposed to find What angle does this vector make with the positive x axis

$tan \theta =\frac{A_y}{A_x}\\tan \theta =\frac{8}{-10}\\\theta =tan^{-1}(\frac{8}{-10})\\\theta =tan^{-1}(\frac{8}{-10})=-38.65^{\circ}$

Now ,

$\alpha = 180-38.65=141.35^{\circ}$

So, this vector make an angle of 141° with the positive x axis

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Combine like terms 2−4y3+5y3+y3+2+5x3+2

Likurg_2 [28]

Answer: 10y3 + 5x3 + 2

Step-by-step explanation:

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3 years ago

A square coffee table has an area of 196 square inches.what is the perimeter

andrew11 [14]

Answer:

56 in

Step-by-step explanation:

Square ....so each side is = x

x * x = area = 196

x^2 = 196

x = sqrt (196) = 14 in

perimeter = x + x + x + x = 56 in

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2 years ago

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Plz help me with this

shtirl [24]

2x-7y=18 _(1)
-2x+4y=5

11y=23 , Y=23/11 sub in 1

2x-7×23/11=18

2x = 18×11 + 7×23

x = ( 198 + 161 )/2 = 359/2

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4 years ago

Factor f(x) = 15x^3 - 15x^2 - 90x completely and determine the exact value(s) of the zero(s) and enter them as a comma separated

Illusion [34]

Answer:

$x=-2,0,3$

Step-by-step explanation:

We have been given a function $f(x)=15x^3-15x^2-90x$ . We are asked to find the zeros of our given function.

To find the zeros of our given function, we will equate our given function by 0 as shown below:

$15x^3-15x^2-90x=0$

Now, we will factor our equation. We can see that all terms of our equation a common factor that is $15x$ .

Upon factoring out $15x$ , we will get:

$15x(x^2-x-6)=0$

Now, we will split the middle term of our equation into parts, whose sum is $-1$ and whose product is $-6$ . We know such two numbers are $-3\text{ and }2$ .

$15x(x^2-3x+2x-6)=0$

$15x((x^2-3x)+(2x-6))=0$

$15x(x(x-3)+2(x-3))=0$

$15x(x-3)(x+2)=0$

Now, we will use zero product property to find the zeros of our given function.

$15x=0\text{ (or) }(x-3)=0\text{ (or) }(x+2)=0$

$15x=0\text{ (or) }x-3=0\text{ (or) }x+2=0$

$\frac{15x}{15}=\frac{0}{15}\text{ (or) }x-3=0\text{ (or) }x+2=0$

$x=0\text{ (or) }x=3\text{ (or) }x=-2$

Therefore, the zeros of our given function are $x=-2,0,3$ .

7 0

4 years ago

Which table represents the graph of a logarithmic function in the form y=log3x when b>1?

alex41 [277]

Answer:

The satisfied table of the given function $y = log_{b} (x)$

x 1/8 1/4 1/2 1 2

y -3 -2 -1 0 1

Step-by-step explanation:

Explanation :-

Given logarithmic function $y = log_{b} (x)$ if b >1

Given first table

put x = $\frac{1}{8}$ given b > 1 so we can choose b = 2

$y = log_{2} (\frac{1}{8} )$

$y = log_{2} (2^{-3} )$

we will apply logarithmic formula

log x ⁿ = n log (x)

$y = log_{2} (2^{-3} ) = -3 log_{2} (2) = -3 (1) = -3$

y = -3

ii)

put x = $\frac{1}{4}$ given b > 1 so we can choose b = 2

$y = log_{2} (\frac{1}{4} )$

$y = log_{2} (2^{-2} )$

we will apply logarithmic formula

log x ⁿ = n log (x)

$y = log_{2} (2^{-2} ) = -2 log_{2} (2) = -2 (1) = -2$

y = -2

iii)

put x = $\frac{1}{2}$ given b > 1 so we can choose b = 2

$y = log_{2} (\frac{1}{2} )$

$y = log_{2} (2^{-1} )$

we will apply logarithmic formula

log x ⁿ = n log (x)

$y = log_{2} (2^{-1} ) = -1 log_{2} (2) = - (1) = -1$

y = -1

iv)

put x = 1 given b > 1 so we can choose b = 2

$y = log_{2} (1 )$ = 0

y = 0

v)

put x = $2$ given b > 1 so we can choose b = 2

$y = log_{2} (2 )$

y = 1

Final answer:-

The satisfied table of the given function

x 1/8 1/4 1/2 1 2

y -3 -2 -1 0 1

8 0

3 years ago

Read 2 more answers