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3 years ago

9

A spinner is divided into four colored sections that are not of equal size: red blue purple and orange Orange : 9 red 15 blue 24

purple 12 based on these results what is the probability that the arrow will land on the purple section

1 answer:

MatroZZZ [7]3 years ago

8 0

Answer:

i think it is 24/60 or 2/5

Step-by-step explanation:

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You purchase 26 "parking hours" that you can use over the next month to park

Nadusha1986 [10]

Answer:

The number of weekday hours are 5.

Step-by-step explanation:

<em>Let the number of weekday hours be "x".</em>

<em>The total week hours are 26, then number of weekend hours is (26-x) .</em>

<em>The charges for weekdays are $2 per hour and for weekends $10 per hour.</em>

The total money spent is $220.

$(2)(x) + (26-x)(10) = 220$

$8x = 40$

x = 5.

Thus weekday hours utilized are 5 hours, and weekend are (26-5) 21 hours.

4 0

3 years ago

Write the trigonometric expression in terms of sine and cosine, and then simplify. cot()/sin()-csc()

OLEGan [10]

Answer:

First, we know that:

cot(x) = cos(x)/sin(x)

csc(x) = 1/sin(x)

I can't know for sure what is the exact equation, so I will assume two cases.

The first case is if the equation is:

$\frac{cot(x)}{sin(x)} - csc(x)$

if we replace cot(x) and csc(x) we get:

$\frac{cot(x)}{sin(x)} - csc(x) = \frac{cos(x)}{sin(x)} \frac{1}{sin(x)} - \frac{1}{sin(x)}$

Now let's we can rewrite this as:

$\frac{cos(x)}{sin(x)} \frac{1}{sin(x)} - \frac{1}{sin(x)} =\frac{cos(x)}{sin^2(x)} - \frac{1}{sin(x)}$

$\frac{cos(x)}{sin^2(x)} - \frac{sin(x)}{sin^2(x)} = \frac{cos(x) - sin(x)}{sin^2(x)}$

We can't simplify it more.

Second case:

If the initial equation was

$\frac{cot(x)}{sin(x) - csc(x)}$

Then if we replace cot(x) and csc(x)

$\frac{cos(x)}{sin(x)}*\frac{1}{sin(x) - 1/sin(x)} = \frac{cos(x)}{sin(x)}*\frac{1}{sin^2(x)/sin(x) - 1/sin(x)}$

This is equal to:

$\frac{cos(x)}{sin(x)}*\frac{sin(x)}{sin^2(x) - 1}$

And we know that:

sin^2(x) + cos^2(x) = 1

Then:

sin^2(x) - 1 = -cos^2(x)

So we can replace that in our equation:

$\frac{cos(x)}{sin(x)}*\frac{sin(x)}{sin^2(x) - 1} = \frac{cos(x)}{sin(x)}*\frac{sin(x)}{-cos^2(x)} = -\frac{cos(x)}{cos^2(x)}*\frac{sin(x)}{sin(x)} = - \frac{1}{cos(x)}$

5 0

3 years ago

Use long division to write

julia-pushkina [17]

Answer:

In this photo the steps are given.

The difference between these answers are that 4/9 is a repetitive decimal and that 5/8 is a finite decimal.

5 0

2 years ago

Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2

NNADVOKAT [17]

$y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}$

Divide both sides by $\dfrac13y^{-2}(x)$ :

$3y^2y'-y^3=xe^x\ln x$

Substitute $v(x)=y(x)^3$ , so that $v'(x)=3y(x)^2y'(x)$ .

$v'-v=xe^x\ln x$

Multiply both sides by $e^{-x}$ :

$e^{-x}v'-e^{-x}v=x\ln x$

The left side can be condensed into the derivative of a product.

$(e^{-x}v)'=x\ln x$

Integrate both sides to get

$e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C$

Solve for $v(x)$ :

$v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

Solve for $y(x)$ :

$y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

$\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}$

4 0

3 years ago

HELP PLEASE ASAP WITH SLOPEE

tamaranim1 [39]

Answer:

y = 1/5x + 7

explain:

rise is one, run is five, intercepts at y=7

5 0

3 years ago