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2 years ago

I need to know what R + 3 =3

Mathematics

2 answers:

navik [9.2K]2 years ago

6 0

Answer:

Step-by-step explanation:

Charra [1.4K]2 years ago

6 0

Answer:

answer:0+3=3

Step-by-step explanation:

to find the answer do the opposite to what is happening to the answer 3-3=0 so 0 is the variable.

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which equations are equivalent to 3/4+m=-7/4 check all that apply m=10/4 m=-10/4 m=-5/2 11/4+m=-1/4 -5/4+m=-15/4 m+2=-0.5 Help f

Alecsey [184]

Answer:

The correct choices are; B,C,E, and F.

Step-by-step explanation:

The given equation is;

$\frac{3}{4}+m=-\frac{7}{4}$

We solve for m to obtain:

$m=-\frac{7}{4}-\frac{3}{4}=-2.5$

We also solve the remaining equations to see which ones give the same result.

A: $m=\frac{10}{4} =2.5$

B: $m=-\frac{10}{4} =-2.5$

C: $m=-\frac{5}{2} =-2.5$

D: $\frac{11}{4}+m=-\frac{1}{4}$

$m=-\frac{1}{4}-\frac{11}{4}=-3$

E: $-\frac{5}{4}+m=-\frac{15}{4}$

$m=-\frac{15}{4}+\frac{5}{4}=-2.5$

F: $m+2=-0.5$

$m=-0.5-2=-2.5$

The equivalent equations are; B,C,E, and F.

3 0

3 years ago

Read 2 more answers

What is the value of h(3) multiplied by h(7)<br><br> P.S. I will give brainliest to the best answer

Eddi Din [679]

Answer:

Step-by-step explanation:

Because h(3) times h(7) is equal to 21.

Hope it helps!

4 0

2 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:

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