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3 years ago

6

-3 1/3-5 2/3 - (-2 1/3)

1 answer:

Leni [432]3 years ago

8 0

Answer:

-20/3

The exact form is -20/3

The decimal form is -6.6 repeating

The mixed number form is -6 2/3

Hope this helps!

<u><em>PLEASE, </em></u>consider brainliest. I only have 3 left and then my rank will go up.

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in the given kite SZ= 10yards , WZ = 10 yards TZ=12 yards , and RZ=32 yards Determine the area of the kite

Fynjy0 [20]

If we assume the given segments are those from the vertices to the point of intersection of the diagonals, it seems one diagonal (SW) is 20 yards long and the other (TR) is 44 yards long. The area (A) of the kite is half the product of the diagonals:

... A = (1/2)·SW·TR = (1/2)·(20 yd)·(44 yd)

... A = 440 yd²

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3 years ago

I need help real quick please!

iogann1982 [59]

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3 years ago

10 - 9 x 5 + 6 x (-2)

olya-2409 [2.1K]

Answer:

The answer to your equation is -47.

Step-by-step explanation:

Follow the order of operations. First do 9*5=45, and 6*-2 to get -12. Next, just finish the equation from left to right to get -47.

5 0

4 years ago

Read 2 more answers

Can someone help me with graphing this function?

Paul [167]

p.s. Excuse the scribble.

Step-by-step explanation:

This is a piece-wise function. The three intervals we need to worry about are [0, 1), [1,2], and [2,4].

Separate the functions into their pieces and draw out the individual graphs. Place them together onto the graph within their respective intervals.

5 0

3 years ago

Find the tangent line equation of the curve at the given point. Y=arcsin(7x) at the point where x=sqrt2/14

Mumz [18]

Answer:

$Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

Step-by-step explanation:

The equation of the curve is

$Y = sin^{-1}(7x)$

To find the equation of tangent we need to differentiate this equation w.r.t x

So, differentiating we get

$Y'=\frac{7}{\sqrt{1-49x^2} }$

This would give the slope of the tangent line at any given point of which x coordinate is known. In the present case it is $x = \sqrt{\frac{1}{7} }$

Then slope would accordingly be

$Y'=\frac{7}{\sqrt{1-49/49} }$

= ∞

For, $x = \sqrt{\frac{1}{7} }$ , $Y = sin^{-1}(7/7)= \pi/2$

Equation of tangent line, in the point slope form, would be $Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

4 0

3 years ago