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3 years ago

25+12 1/2 + 61/4+31/8+19/16+25/32=

1 answer:

4 0

Make them all have the same denominator
32

25+12 16/32+488/32+124/32+38/32+25/32

Add all of the numerators... to get 691/32+1184/32
1875/32
Simplify to get 58 19/32

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the width of a rectangle is the length minus 2 units. The area of the rectangle is48 units. What is the width, in units, of the

tatuchka [14]

Answer:

Step-by-step explanation:

put unknown as y

area of rectangle = width × long

y(y - 2)=48

y² -2y =48

u can use scientific calculator to find this

y²-2y-48=0

(y-8)(y+6)=0

y=8 y=-6

take the positive value and ignore the negative value

width=y-2 long=8

8-2=6

I hope this help you :)

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3 years ago

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Answer:

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3 years ago

Enter the equation of the line in slope-intercept form. Enter the answer in fraction form.

Montano1993 [528]

Answer:

11/2??????

Step-by-step explanation:

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3 years ago

A projectile is launched from the ground at an angle of theta above the horizontal with an initial speed of v in ft/s. The rang

VARVARA [1.3K]

Answer:

$\theta=24.4^\circ$

Step-by-step explanation:

The formula you wrote has the fraction the other way around. You can find that the range of a projectile is in reality approximated by the equation $x=\frac{v^2}{g} sin(2\theta)=\frac{v^2}{32ft/s^2} sin(2\theta)$ , where we will use ft for distances.

From the given equation we have then $\frac{x32ft/s^2}{v^2} =sin(2\theta)$ , which means $\theta=\frac{Arcsin(\frac{x32ft/s^2}{v^2})}{2}$ since the arcsin is the inverse function of the sin.

Since we have x = 170 ft and v = 85 ft/s, we can substitute these values from the equation written, and we will have $\theta=\frac{Arcsin(\frac{(170ft)(32ft/s^2)}{(85ft/s)^2})}{2}$ , and from now on we have just to use a calculator, obtaining $\theta=\frac{Arcsin(0.75294117647)}{2}=\frac{48.84579804^\circ}{2}=24.4^\circ$

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3 years ago

Go to Canvas to watch an instructional video and complete the assignment. After watching the video and uploading the assignment

Reptile [31]

This is simply what I wrote for my assignment, sorry I am a bit late.

<em>It was confusing when she took out the negative and postive pairs, but then I soon knew that the negative tiles are like subtracting positive tiles. Then once she got the total I was amazed with how easy it was to find out the value of x.</em>

hope this helps! <3

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2 years ago