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goldfiish [28.3K]

3 years ago

8

Does anyone know if this is correct or how to do this? (part a,b, and c)

1 answer:

stepladder [879]3 years ago

5 0

Part A you would use 200-L ( since you have 400 feet total 200 feet would equal length plus width)

Part B 200 - 80 = 120

Part C 200-90 = 110

area = 90 x 110 = 9900 square feet

You might be interested in

A dolphin jumps from the water at a initial velocity of 16 feet per second the equation h=-8t^2 + 16t models the dolphins height

german

Answer:

The answer to the equation -8t^2+16t = -48

Step-by-step explanation:

Add ''-8t^2'' to ''16t'' and the answer you get is ''-48''

6 0

3 years ago

Read 2 more answers

Of the three properties, reflexive, symmetric, and transitive that define the relation "is equal to," which one could also apply

yan [13]

Answer: Transitive property.

Step-by-step explanation:

First, for the equality we have:

Reflexive:

For all real numbers x, x = x.

Symmetric:

For all real numbers x, y

if x= y, then y = x.

Transitive:

For reals x, y and z.

if x = y, and y = z, then x = z.

Now, let's talk about inequalities.

first, the reflexive property will say that:

x > x.

This has no sense, so this property does not work for inequalities.

Now, the reflexive.

If x > y, then y > x.

Again, this has no sense, if x is larger than y, then we can never have that y is larger than x. This property does not work for inequalities.

Not, the transitive property.

if x > y, and y > z, then x > z.

This is true.

x is bigger than y, and y is bigger than z, then x should also be bigger than z.

x > y > z.

And this also works for the inverse case:

x < y and y < z, then x < z.

So the correct option is transitive property.

3 0

3 years ago

Plot the point whose polar coordinates are given. Then find two other pairs of polar coordinates of this point, one with r >

Ira Lisetskai [31]

Answer:

The other pairs are:

$(a)\ (2, \frac{5\pi}{6}) \to$ $(2, \frac{17\pi}{6})$ and $(-2, \frac{23\pi}{6})$

$(b)\ (1, -\frac{2\pi}{3}) \to$ $(1, \frac{4\pi}{3})$ and $(-1, \frac{7\pi}{3})$

$(c)\ (-1, \frac{5\pi}{4}) \to$ $(-1, \frac{3\pi}{4} )$ and $(1, \frac{7\pi}{4})$

See attachment for plots

Step-by-step explanation:

Given

$(a)\ (2, \frac{5\pi}{6})$

$(b)\ (1, -\frac{2\pi}{3})$

$(c)\ (-1, \frac{5\pi}{4})$

Solving (a): Plot a, b and c

See attachment for plots

Solving (b): Find other pairs for $r > 0$ and $r < 0$

The general rule is that:

The other points can be derived using

$(r, \theta) = (r, \theta + 2n\pi)$

and

$(r, \theta) = (-r, \theta + (2n + 1)\pi)$

Let $n =1$ ---- You can assume any value of n

So, we have:

$(r, \theta) = (r, \theta + 2n\pi)$

$(r, \theta) = (r, \theta + 2*1*\pi)$

$(r, \theta) = (r, \theta + 2\pi)$

$(r, \theta) = (-r, \theta + (2n + 1)\pi)$

$(r, \theta) = (-r, \theta + (2*1 + 1)\pi)$

$(r, \theta) = (-r, \theta + (2 + 1)\pi)$

$(r, \theta) = (-r, \theta + 3\pi)$

$(a)\ (2, \frac{5\pi}{6})$

$r = 2\ \ \ \ \theta = \frac{5\pi}{6}$

So, the pairs are:

$(r, \theta) = (r, \theta + 2\pi)$

$(2, \frac{5\pi}{6}) = (2, \frac{5\pi}{6} + 2\pi)$

Take LCM

$(2, \frac{5\pi}{6}) = (2, \frac{5\pi+12\pi}{6})$

$(2, \frac{5\pi}{6}) = (2, \frac{17\pi}{6})$

And

$(r, \theta) = (-r, \theta + 3\pi)$

$(2, \frac{5\pi}{6}) = (-2, \frac{5\pi}{6} + 3\pi)$

Take LCM

$(2, \frac{5\pi}{6}) = (-2, \frac{5\pi+18\pi}{6})$

$(2, \frac{5\pi}{6}) = (-2, \frac{23\pi}{6})$

The other pairs are:

$(2, \frac{17\pi}{6})$ and $(-2, \frac{23\pi}{6})$

$(b)\ (1, -\frac{2\pi}{3})$

$r = 1\ \ \ \theta = -\frac{2\pi}{3}$

So, the pairs are:

$(r, \theta) = (r, \theta + 2\pi)$

$(1, -\frac{2\pi}{3}) = (1, -\frac{2\pi}{3} + 2\pi)$

Take LCM

$(1, -\frac{2\pi}{3}) = (1, \frac{-2\pi+6\pi}{3})$

$(1, -\frac{2\pi}{3}) = (1, \frac{4\pi}{3})$

And

$(r, \theta) = (-r, \theta + 3\pi)$

$(1, -\frac{2\pi}{3}) = (-1, -\frac{2\pi}{3} + 3\pi)$

Take LCM

$(1, -\frac{2\pi}{3}) = (-1, \frac{-2\pi+9\pi}{3})$

$(1, -\frac{2\pi}{3}) = (-1, \frac{7\pi}{3})$

The other pairs are:

$(1, \frac{4\pi}{3})$ and $(-1, \frac{7\pi}{3})$

$(c)\ (-1, \frac{5\pi}{4})$

$r = -1 \ \ \ \ \theta = \frac{-5\pi}{4}$

So, the pairs are

$(r, \theta) = (r, \theta + 2\pi)$

$(-1, \frac{-5\pi}{4}) = (-1, \frac{-5\pi}{4} + 2\pi)$

Take LCM

$(-1, \frac{-5\pi}{4}) = (-1, \frac{-5\pi+8\pi}{4} )$

$(-1, \frac{-5\pi}{4}) = (-1, \frac{3\pi}{4} )$

And

$(r, \theta) = (-r, \theta + 3\pi)$

$(-1, \frac{-5\pi}{4}) = (-(-1), \frac{-5\pi}{4}+ 3\pi)$

Take LCM

$(-1, \frac{-5\pi}{4}) = (1, \frac{-5\pi+12\pi}{4})$

$(-1, \frac{-5\pi}{4}) = (1, \frac{7\pi}{4})$

So, the other pairs are:

$(-1, \frac{3\pi}{4} )$ and $(1, \frac{7\pi}{4})$

5 0

3 years ago

Find the value of x.

Crank

Answer: $x=76\°$

Step-by-step explanation:

By definition the angle formed by the intersection of two chords inside a circle is:

$Angle\ formed\ by\ two\ Chords=\frac{sum\ of\ intercepted\ Arcs}{2}$

You can observe in the figure that the angle formed by the intersection of the two chords measures 94°, and the intercepteed arcs are: x° and 112°.

Therefore, you need to substitute these values into the formula and solve for "x to find its value. Then:

$94\°=\frac{x+112\°}{2}\\\\(94\°)(2)=x+112\°\\\\188\°-112\°=x\\\\x=76\°$

5 0

3 years ago

K is no greater than 9?<br><br> Write>,<.

alekssr [168]

Greater than symbol is ">"
But since he mentioned "NO greater than" then it's the opposite of greater than, which means "K" is less than "9"

K<9

4 0

3 years ago