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marishachu [46]

3 years ago

12

Need help plsssssss help Here are two number machines, A and B.

2 answers:

olga2289 [7]3 years ago

5 0

Answer:

Step-by-step explanation:so I think

You +8×5=40

-4×3=-12

hoa [83]3 years ago

5 0

Answer:

the answer is x=-7

Step-by-step explanation:

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Evaluate the expression when<br> c+5

expeople1 [14]

Answer:

what is the expression

Step-by-step explanation:

7 0

3 years ago

-3y+11=-8y+51 <br> What is y? Simplify your answer as much as possible

icang [17]

Y=8 ........................

3 0

3 years ago

Read 2 more answers

Phyllis invested $11,500, a portion earning a simple interest rate of

Cerrena [4.2K]

Answer:

He invested $8500 with rate 3 1/5%

He invested $3000 with rate 3%

Step-by-step explanation:

* <em>Lets revise the rule of the simple interest</em>

- Simple interest I = PRT, where P in the money invested, R is the

interest rate and T is the time of investment

- Phyllis invested $11,500

- A portion earning a simple interest rate of 3 1/5% (3.2%) per year

- The rest earning a rate of 3% per year

- After one year the total interest earned on these investments

was $362.00

- We need to find the amount of money invested in each rate

* Assume that he invested $x with rate 3 1/5% and $y with rate 3%

∵ He invested $11,500 in both rates

∴ x + y = 11,500 ⇒ (1)

∵ He invested $x with 3 1/5%

∵ P = $x

∵ R = 3.2/100 = 0.032

∵ T = 1

∴ $I_{1}=x(0.032)(1)$

∴ $I_{1}=0.032x$

∵ He invested $y with 3%

∵ P = $y

∵ R = 3/100 = 0.03

∵ T = 1

∴ $I_{2}=y(0.03)(1)$

∴ $I_{2}=0.03y$

∵ The total interest was $362.00

- The total interest = $I_{1}$ + $I_{2}$

∴ 0.032x + 0.03y = 362.00 ⇒ (2)

<em>We have system of equations</em>

x + y = 11,500 ⇒ (1)

0.032x + 0.03y = 362.00 ⇒ (2)

Multiply equation (1) by -0.03 to eliminate y

-0.03x - 0.03y = 345 ⇒ (3)

Add equations (2) and (3)

0.002x = 17

Divide both sides by 0.002

x = 8500

Substitute the value of x in equation (1)

8500 + y = 11,500

Subtract 8500 from both sides

y = 3000

* He invested $8500 with rate 3 1/5%

* He invested $3000 with rate 3%

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Chuge%20%5Cpink%7Bwhat%20%5C%3A%20is%20%5C%3A%20cranial%7D%20%5C%5C%20%5Chuge%20%5Cpink%7

worty [1.4K]

$\huge\mathcal{\fcolorbox{aqua}{pink}{\blue{✿Answer ❖}}}$

“Nerves that extend throughout the body on both sides emerging directly from brain and brain stem are called cranial nerves.”

6 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