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SVETLANKA909090 [29]
3 years ago
13

Math :) :) I need HelPpPPpPPPP lol

Mathematics
1 answer:
11111nata11111 [884]3 years ago
6 0

Answer:

27b^6

Step-by-step explanation:

3^3 is 27

(b^2)^3 is b^6

So the answer is A. 27b^6

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How many three-digit positive integers <img src="https://tex.z-dn.net/?f=x" id="TexFormula1" title="x" alt="x" align="absmiddle"
Anastaziya [24]

Answer:

  40

Step-by-step explanation:

Any solution x will mod 23 will also have x+23n as a solution, for some integer n. Since 900/23 = 39 3/23, we know there are 39 or 40 three-digit integers of this form.

As it happens, 100 is the smallest 3-digit solution. So, there are 40 three-digit numbers that are of the form 100 +23n, hence 40 solutions to the equation.

_____

The equation reduces, mod 23, to ...

  10x = 11

Its solutions are x = 23n +8.

3 0
3 years ago
Sinx=cos19∘<br> What is the value of x?<br><br> Enter your answer in the box.
rusak2 [61]
Remember that cos is just a translation of sin and vice versa.
So:
Sin(x) = Cos(90 - x)
Cos(x) = Sin(90 - x)

Therefore, to answer your question:
Cos(19) = Sin(90 - 19)
= Sin(71)
6 0
3 years ago
Plsss help me !!!!
Yuri [45]

Answer:

The enrollment after 5 years is 10,724

Step-by-step explanation:

Generally, we can have the depreciation formula written as follows;

A = P(1 - r)^t

A is the number of enrollment in after a certain number of years t

P is the initial population which is 13,500

r is the rate of depreciation which is 4.5% = 4.5/100 = 0.045

t = 5 years

Substituting these values, we have it that;

A = 13,500(1-0.045)^5

A = 10,723.84

4 0
3 years ago
And also this question as well
gtnhenbr [62]

Answer:

It's the second one I believe, because it's not a straight line, and nearly all functions have curved lines

Step-by-step explanation:

4 0
3 years ago
Find the six trig function values of the angle 240*Show all work, do not use calculator
-BARSIC- [3]

Solution:

Given:

240^0

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

sin240^0=sin(180+60)

Using the trigonometric identity;

sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny

Hence,

\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\  \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\  \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

cos240^0=cos(180+60)

Using the trigonometric identity;

cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny

Hence,

\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\  \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\  \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

tan240^0=tan(180+60)

Using the trigonometric identity;

tan(180+x)=tan\text{ }x

Hence,

\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\  \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}

To get cosec 240 degrees:

\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\  \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\  \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\  \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}

To get sec 240 degrees:

\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\  \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\  \\ Thus, \\ sec240^0=-2 \end{gathered}

To get cot 240 degrees:

\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\  \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\  \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\  \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}

5 0
1 year ago
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