Someone please help me with this

What is the difference of the whole number and the mixed number below? 8-2 7/8 A. 6 7/8 B. 6 1/8 C. 5 7/8 D. 5 1/8

Mariana [72]

Answer:

D. 5 1/8

Step-by-step explanation:

First, convert the mixed number into an improper fraction

8-2 7/8 = 8- 23/8

Calculate the difference

8-23/8 = 41/8

Simplify the fraction

41/8 = 5 1/8

7 0

3 years ago

A 140 kg camera is suspended by two wires over a 40 metre wide football field to get shots of the action from above. At one poin

Katen [24]

Answer:

Please red the answer below

Step-by-step explanation:

In order to determine the length of each cable you use the Newton second law for each component of the forces involved in the situation.

For the x component you have:

$T_1cos\theta_1-T_2cos\theta_2=0$ (1)

T1: tension of the first cable = 1500N

T2: tension of the second cable = 800N

θ1: angle between the horizontal and the first cable

θ2: angle between the horizontal and the first cable

For the y component you have:

$T_1sin\theta_1+T_2sin\theta_2-W=0$ (2)

W: weight of the camera = Mg = (140kg)(9.8m/s^2) = 1372N

You can squared both equations (1) and (2) and the sum the two equations:

$T_1^2cos^2\theta_1=T_2^2cos^2\theta_2\\\\T_1^2sin^2\theta_1=T_2^2sin^2\theta_2-2WT_2sin\theta_2+W^2$

Then, you sum the equations:

$T_1^2(cos^2\theta_1+sin^2\theta_1)=T_2^2(sin^2\theta_2+cos^2\theta_2)-2Wsin\theta_2+W^2$ (3)

Next, you use the following identity:

$sin^2\theta+cos^2\theta=1$

and you obtain in the equation (3):

$T_1^2=T_2^2-2WT_2sin\theta_2+W^2\\\\sin\theta_2=\frac{T_2^2-T_1^2+W^2}{2WT_2}=\frac{(800N)^2-(1500)^2+(1372N)^2}{2(800N)(1372N)}=0.066\\\\\theta_2=sin^{-1}0.066=27.23\°$

With this values you can calculate the value of the another angle, by using the equation (1):

$\theta_1=cos^{-1}(\frac{T_2cos(27.23\°)}{T_1})=cos^{-1}(\frac{(800N)(cos27.23\°)}{1500N})\\\\\theta_1=61.69\°$

Now, you can calculate the length of each cable by using the information about the width of the football field. You use the following trigonometric relation:

$l_1cos\theta_1=40-d\\\\l_2cos\theta_2=d\\\\$

d: distance to the right side of the field

By using the cosine law you can fins a system of equation and then you can calculate the values of l1 and l2.

3 0

3 years ago

write in slope intercept form the equasion of the line that is perpendicular to the line y=-1/2x+1 and passes through the point

cricket20 [7]

Answer:

y = 2x - 11

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - $\frac{1}{2}$ x + 1 ← is in slope- intercept form

with slope m = - $\frac{1}{2}$

Given a line with slope m then the slope of a line perpendicular to it is

$m_{perpendicular}$ = - $\frac{1}{m}$ = - $\frac{1}{-\frac{1}{2} }$ = 2 , then

y = 2x + c ← is the partial equation

To find c substitute (8, 5) into the partial equation

5 = 16 + c ⇒ c = 5 - 16 = - 11

y = 2x - 11 ← equation of perpendicular line

8 0

3 years ago

Find an equation for the inverse of the function. f(x)=2x-3x/4

Drupady [299]

Answer:

f⁻¹(x) = (4x)/5

Step-by-step explanation:

f(x) = 2x - 3x/4.

Assume y=f(x). Therefore:

y = 2x - 3x/4.

Make x the subject in the above equation.

y = (2x(4) - 3x)/4.

y = 5x/4.

4y = 5x.

(4y)/5 = x.

x = (4y)/5.

Therefore:

f(y) = (4y)/5.

Replace y with x.

f⁻¹(x) = (4x)/5.

Therefore, the inverse of f(x) = 2x - 3x/4 is f⁻¹(x) = (4x)/5!!!

7 0

4 years ago

Find the inverse of the function. f(x) = the cube root of quantity x divided by eight. - 4

Lapatulllka [165]

$\bf f(x)=y=\sqrt[3]{\cfrac{x}{8}}-4\impliedby \textit{first, let's switch the variables about}
\\\\\\
\underline{x}=\sqrt[3]{\cfrac{\underline{y}}{8}}-4\implies x+4=\sqrt[3]{\cfrac{\underline{y}}{8}}\implies (x+4)^3=\cfrac{y}{8}
\\\\\\
\boxed{8(x+4)^3=y}\impliedby f^{-1}(x)$

7 0

3 years ago

Read 2 more answers

Someone please help me with this​

Someone please help me with this