All categories

4 years ago

What is the soulution of -(-5-6x) = 4(5x+3)?

1 answer:

4 0

You have to distribute the parentheses first. Since there is a negative sign in front of the parentheses, it counts as a negative 1. After you distribute you should get: -5+6x=20x+12, but that is not your final answer. After you add like terms: 17x=-17. Divide 17 on both sides: x=-1. X=-1 is your final answer.

You might be interested in

If a right triangle is rotated around the longest leg what three-dimensional figure is formed

pantera1 [17]

I think the answer is a cone

4 0

3 years ago

If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe

Natalka [10]

Answer:

They encircle the planet $3.76\times 10^{11}$ times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is $6.022\times 10^{23}$

Thus, the number of dust particles is:

$2.5\times 6.022\times 10^{23}=15.055\times 10^{23}$

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: $10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m$

If we line up the particles the distance they could cover is:

$15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}$

Circumference in meters:

$40,076km\times \frac{1000m}{1km}=40,076,000 m$

Therefore,

$\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}$

Hence, they encircle the planet $3.76\times 10^{11}$ times.

8 0

3 years ago

What is the median of the dot plot for 4,6,6,7,9

jeka57 [31]

Answer:

Step-by-step explanation:

6 is in the middle with 2 numbers on each side so it is the medium

8 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

Hippalectryon

horrorfan [7]

Since the stadium is 6 mi from the practice field which is itself 1 mi from the house, it can either be (1+6)=7 mi North or (1-6)=-5 -> 5 mi South depending on the North/South direction.

the answer is thus A

7 0

3 years ago