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2 years ago

Tan x + sqrt(3) = - 2 tan x

Mathematics

1 answer:

Svet_ta [14]2 years ago

8 0

9514 1404 393

Answer:

nπ -π/6 . . . for any integer n

Step-by-step explanation:

tan(x) +√3 = -2tan(x) . . . . . given

3tan(x) = -√3 . . . . . . . . . . . add 2tan(x)-√3

tan(x) = -√3/3 . . . . . . . . . . divide by 3

x = arctan(-√3/3) = -π/6 . . . . use the inverse tangent function to find x

This is the value in the range (-π/2, π/2). The tangent function repeats with period π, so the set of values of x that will satisfy this equation is ...

x = n·π -π/6 . . . . for any integer n

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Find the x-intercepts of the parabola with vertex (-5,80) and y-intercept (0,-45). Write your answer in this form (x of 1 , y of

Ivan

A parabola is a quadratic function, and a quadratic can be expressed in vertex form, which is:

y=a(x-h)^2+k, where (h,k) is the vertex (absolute maximum or minimum point of the quadratic)

In this case we are given that (h,k) is (-5,80) so we have so far:

y=a(x--5)^2+80

y=a(x+5)^2+80, we are also told that it passes through the point (0,-45) so:

-45=a(0+5)^2+80

-45=25a+80 subtract 80 from both sides

-125=25a divide both sides by 25

-5=a, so now we know the complete vertex form is:

y=-5(x+5)^2+80

The x-intercepts occur when y=0 so:

0=-5(x+5)^2+80 add 5(x+5)^2 to both sides

5(x+5)^2=80 divide both sides by 5

(x+5)^2=16 take the square root of both sides

x+5=±√16 which is

x+5=±4 subtract 5 from both sides

x=-5±4 so the x-intercepts are:

x=-1 and -9

7 0

3 years ago

What is the y-value when x equals 7?<br> y = 310 - 25(x)

monitta

Answer:

y=135

Step-by-step explanation:

To find the solution, we must put 7 in for x and solve, as shown:

$y=310-(25)(7)\\y=310-175\\y=135$

6 0

3 years ago

Read 2 more answers

A triangular prism and two nets are shown:

leonid [27]

The net for a triangular prism consists of
2 identical triangles, here all nets have sides 5,12, and 13.
3 rectangles, all with a common dimension: the height H=14 of the prism.

Each of the 3 rectangles should have dimensions
H=14 plus one of the following:
each of 5, 12, 13 corresponding to one side of the base (triangle).

So the dimensions of the rectangles are
5x14, 12x14, 13x14

And dimensions of the triangles are
5,12 and 13.

The total surface area is therefore
(5+12+13)*14 + 2*(5*12/2)
=420+60
=480.

There is only one net that satisfies this condition.

8 0

3 years ago

Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. r = -5 - 5 cos θ

Grace [21]

Y-axis symmetry=(r, theta)=(-r,theta)
-5-5cos(theta)=r
-r=5+5cos(theta)
no y-axis symmetry
x-axis symmetry=(r,theta)=(r,-theta)
cosine is an even function, so yes it is symmetric around x-axis
origin symmetry=(r,theta)=(-r,theta) or (r, theta+pi)
no, as there is no y-axis symmetry

6 0

3 years ago

April worked 1 1/2 times as long on her math project as did Carl. Debbie worked 1 1/4 times as long as Sonia. Richard worked 1 3

vlada-n [284]

Answer:

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

Step-by-step explanation:

Some data's were missing so we have attached the complete information in the attachment.

Given:

Number of Hours Carl worked on Math project = $5\frac{1}{4}\ hrs$

$5\frac{1}{4}\ hrs$ can be Rewritten as $\frac{21}{4}\ hrs$

Number of Hours Carl worked on Math project = $\frac{21}{4}\ hrs$

Number of Hours Sonia worked on Math project = $6\frac{1}{2}\ hrs$

$6\frac{1}{2}\ hrs$ can be rewritten as $\frac{13}{2}\ hrs$

Number of Hours Sonia worked on Math project = $\frac{13}{2}\ hrs$

Number of Hours Tony worked on Math project = $5\frac{2}{3}\ hrs$

$5\frac{2}{3}\ hrs$ can be rewritten as $\frac{17}{3}\ hrs.$

Number of Hours Tony worked on Math project = $\frac{17}{3}\ hrs.$

Now Given:

April worked $1\frac{1}{2}$ times as long on her math project as did Carl.

$1\frac{1}{2}$ can be Rewritten as $\frac{3}{2}$

Number of Hours April worked on math project = $\frac{3}{2}$ $\times$ Number of Hours Carl worked on Math project

Number of Hours April worked on math project = $\frac{3}{2}\times \frac{21}{4} = \frac{63}{8}\ hrs \ \ Or \ \ 7\frac{7}{8} \ hrs$

Also Given:

Debbie worked $1\frac{1}{4}$ times as long as Sonia.

$1\frac{1}{4}$ can be Rewritten as $\frac{5}{4}$ .

Number of Hours Debbie worked on math project = $\frac{5}{4}$ $\times$ Number of Hours Sonia worked on Math project

Number of Hours Debbie worked on math project = $\frac{5}{4}\times \frac{13}{2}= \frac{65}{8}\ hrs \ \ Or \ \ 8\frac{1}{8}\ hrs$

Also Given:

Richard worked $1\frac{3}{8}$ times as long as tony.

$1\frac{3}{8}$ can be Rewritten as $\frac{11}{8}$

Number of Hours Richard worked on math project = $\frac{11}{8}$ $\times$ Number of Hours Tony worked on Math project

Number of Hours Debbie worked on math project = $\frac{11}{8}\times \frac{17}{3}= \frac{187}{24}\ hrs \ \ Or \ \ 7\frac{19}{24}\ hrs$

Hence We will match each student with number of hours she worked.

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

5 0

3 years ago

Read 2 more answers