Solve tan^2 theta + 1 = -2 tan theta on the interval 0° ≤ θ ≤ 360°.

Answer:

The answer is C 3- sqrt 11

Step-by-step explanation:

Answer:

2+3+10+15+20+2+30=82

82/8=10.25

Step-by-step explanation:+10.25

Answer:

The answer is 14.9

Step-by-step explanation:

First I added 2,3,10,15,20,24,30 then  divided it by 7.

Answer:

Step-by-step explanation:

We will use the work form of a quadratic to determine what a is...in fact we will write the equation for the whole thing in the process, because it's part of solving for a.

y = ±|a|(x - h)² + k

where x and y are from a coordinate point on the graph, h and k are the coordinates of the vertex, the absolute value of a indicates how steep or flat the graph is compared to the parent graph, and the ± is because a positive parabola opens up and a negative one opens upside down.

The vertex is (0, 9) and the coordinate point I chose to use is (3, 0). Filling those in and solving for a:

0 = ±|a|(3 - 0)² + 9 and

0 = ±|a|(3)² + 9 and

-9 = ±|a|9 and

-1 = ±|a| so a = 1. Because this is an upside down parabola the negative is out front, but a is independent of it. The correct choice is C. The quadratic function is

[tex]y=-x^2+9[/tex] or in more detailed form:

[tex]y=-(x-0)^2+9[/tex]

Answer:

-2

Step-by-step explanation:

Answer:

Option (2).

Step-by-step explanation:

This question is incomplete; here is the complete question and find the figure attached.

In the diagram of a circle O, what is the measure of ∠ABC?

27°

54°

108°

120°

m(minor arc [tex]\widehat {AC}[/tex]) = 126°

m(major arc [tex]\widehat {AC}[/tex]) = 234°

By the intersecting tangents theorem,

If the two tangents of a circle intersect each other outside the circle, measure of angle formed between them is half the difference of the measures of the intercepted arcs.

m∠ABC = [tex]\frac{1}{2}[m(\text {major arc{AC})}-m(\text{minor arc} {AC})][/tex]

             = [tex]\frac{1}{2}(234-126)[/tex]

             = 54°

Therefore, Option (2) will be the answer.

Answer:

B.

Step-by-step explanation:

f(x)=x²−4x−21

The degree is the biggest power of x.  That's a polynomial of degree 2, also called a quadratic function.  Let's find its zeros.

0 = x²−4x−21 = (x - 7)(x+3)

x=7 or x=-3

The fundamental theorem guarantees every non-constant polynomial with complex coefficients has a complex zero, let's call it r.  If we divide the polynomial by x-r there won't be any remainder and we'll get a new polynomial, one degree less.  The fundamental theorem again applies and (if it's not a constant polynomial) we are assured of another zero, s.  We divide by x-s and get a new polynomial of degree one less.  We repeat all this until we get a constant polynomial (degree zero).    So we get a zero for every degree. They're not necessarily all different.

Answer:

The degree of f(x) is 2. The Fundamental Theorem of Algebra guarantees that a polynomial equation has the same number of complex roots as its degree. This means that f(x) has exactly 2 zeros. Those zeros are 7 and -3.

Answer:a

Step-by-step explanation:

Answer:

The answer is A.

Step-by-step explanation:

Pythagorean theorem.

a^2+b^2=c^2.

2^2+3^2=d^2

Answer:

hope this helps you

Answer:

4

Step-by-step explanation:

g(3) = -2(3)^2 -4 = 2(9) -4 = 18-4 = 14

|f(2)| = | 2^2 -3| = | 4-3| = 1

(g(3) +2) = (14+2) = 16

4*  |f(2)|  = 4*1 = 4

16/4 = 4

Answer:

C. (1,-1)

Step-by-step explanation:

Lines are intersecting each other at point (1, - 1).

Hence, the system of equations graphed here would be (1, - 1)

Answer:1.8*10^-1

Step-by-step explanation:

Answer:     1.8 x 10^−1

Step-by-step explanation:

Answer:

There is one solution

Step-by-step explanation:

4x + 6 = 6x - 2

Subtract 4x from each side

4x-4x + 6 = 6x-4x - 2

6 = 2x-2

Add 2

6+2 = 2x

8 = 2x

Divide by 2

8/2 = 2x/2

4=x

7

tan(60) = h/40

h = 40 x tan(60)

h = 40√3

cos(60) = 40/r

r = 40/cos(60)

r = 80

8.

tan(60) = p/82

p = 82 x tan(60)

p = 82√3

cos(60) = 82/t

t = 82/cos(60)

t = 164

9.

tan(60) = 22√3 /x

x = 22√3 /tan(60)

x = 22

sin(60) =  22√3 /k

k = 22√3 / sin(60)

k = (20/3)√3

Answer:

12 square units

Step-by-step explanation:

formula= a+b/2×h

a=2

b=6

h=3

the answer will be = 2+6/2×3

=8/2×3

=4×3

=12

Answer:

17 square units i think

Answer:

When we divide the inequality by 3 we get x < -2 so the answer is A.

Answer:

The fourth choice is the correct one.

Step-by-step explanation:

If the count of negative signs is even, the product is even (positive).

This does not apply to the first choice; this expression is odd.

Same for the second choice.

Same for the third choice.  The expression is odd.

The fourth choice is POSITIVE because there are an even number (2) of negative signs.

Answer:

To eliminate y- terms

Multiply equ(1) by 3 and equ(2) by -4

To eliminate y- terms

Multiply (1) by 3 and (2) by -5

Step-by-step explanation:

5x - 4y = 28 (1)

3x - 3y = 30 (2)

To eliminate y- terms

Multiply equ(1) by 3 and equ(2) by -4

15x-12y=84

-12x+12y= -120

Add equ (3) and (4)

3x=84-120

3x=-36

x=-36/3

= -12

x=-12

To eliminate x- terms

5x-4y=28

3x-3y=30

Multiply (1) by 3 and (2) by -5

15x-12y=84

-15x+15y=-150

Add the new equation

3y= -66

y= -22

Answer:

c = 20d + 15

Step-by-step explanation:

answer choices may help if the answer is not one of the choices; but here is the answer that it should be:

20 dollars per day is 20 times the length of rental.

15 is a cost that is added and does not change.

c = the total cost therefore:

c = 20d + 15

Step-by-step explanation:

8 + 4x > 5x + 5

8 - 5 > 5x - 4x

3 > x

The solution to the system of equation is x < 3

Given the inequality expression 2(4+2x)>5x+5

Expand the inequality

2(4+2x)>5x+5

8 + 4x > 5x + 5

Collect the like terms

4x - 5x > 5 -8

-x > -3

x < 3

Hence the solution to the system of equation is x < 3

Learn more on inequality here: https://brainly.com/question/24372553

Answer:

-3n⁹

Step-by-step explanation:

= [tex]\sqrt[3]{-27n^(27)}[/tex]

= [tex]\sqrt[3]{-27} * \sqrt[3]{n^{27}}[/tex]

= (-3) × n⁹

= -3n⁹

Answer:

18 square meters

Step-by-step explanation:

Answer:

solution,

The given figure if a Trapezium whose parallel sides are of 3 m and 2 m respectively.

Distance between the parallels sides

i.e. height is 3 m

Now,

[tex]area = \frac{1}{2} \times sum \: of \: parallel \: sides \times height \\ = \frac{1}{2} \times (3 + 2) \times 3 \\ = \frac{1}{2} \times 5 \times 3 \\ = \frac{15}{2} \\ = 7.5 \: {m}^{2} [/tex]

Hope this helps...

Good luck on your assignment...

Answer:

Ryan had a head start of 10 meters

Answer:

Ryan had a head start of 10 meters.

Have a great day!

Step-by-step explanation:

Please mark me brainliest!

Answer:

[tex]-y^2+6y+16[/tex]

Step-by-step explanation:

[tex]-(y+2)(y-8)[/tex]

[tex]-1(y+2) \times (y-8)[/tex]

[tex](-y-2)\times (y-8)[/tex]

[tex]-y(y-8)-2(y-8)[/tex]

[tex]-y^2+8y-2y+16[/tex]

[tex]-y^2+6y+16[/tex]

Step-by-step explanation:

a 6xy is the answer to question a

Answer:

(1) 6 (2) 2

Step-by-step explanation:

[tex]1)y=\frac{36}{6}\\y=62) \\[/tex]

[tex]y=\frac{36}{18}\\y=2[/tex]

Answer:

The value of x is [tex]x=\frac{\log(2)}{\log(6)}[/tex].

Step-by-step explanation:

Solve the equation as follows:

[tex]4\times 6^{x}-7=1[/tex]

     [tex]4\times 6^{x}=7+1[/tex]

           [tex]6^{x}=\frac{8}{4}[/tex]

           [tex]6^{x}=2[/tex]

Take log on both sides.

[tex]\log(6^{x})=\log(2)[/tex]

[tex]x\log (6)=\log(2)[/tex]

        [tex]x=\frac{\log(2)}{\log(6)}[/tex]

Thus, the value of x is [tex]x=\frac{\log(2)}{\log(6)}[/tex].

Answer:

[tex]r=\frac{44}{\pi }[/tex]

Step-by-step explanation:

[tex]88=2\pi r[/tex]

[tex]\mathrm{Switch\:sides}\\2\pi r=88\\\frac{2\pi r}{2\pi }=\frac{88}{2\pi }\\r=\frac{44}{\pi }\\or\\r=14.00563[/tex]

Answer:

2 metres

Step-by-step explanation:

Circumference = 2 × π × r

→ Substitute in the values

88 = 2 ×  [tex]\frac{22}{7}[/tex]  × r

→ Divide both sides by 2 to isolate  [tex]\frac{22}{7}[/tex]  and r

44 =  [tex]\frac{22}{7}[/tex]  × r

→ Multiply everything by 7 to get rid of the fraction

308 = 22 × 7r

→ Divide the equation by 22 to isolate 7r

14 = 7r

→ Divide the equation by 7 to isolate r

2 = r

The radius of the circle with a circumference of 88 meters is 2 metres

The right answer is 7/3

please see the attached picture for full solution

hope it helps..

Good luck on your assignment

Answer:

[tex]slope = \frac{7}{3} [/tex]

Step-by-step explanation:

[tex](5 \: \: , \: \: 9) = > (x1 \: \: \:, \: y1) \\ (2 \: \: , \: \: 2) = > (x2 \: \: \:, \: \: y2)[/tex]

[tex] slope \\ = \frac{y1 - y2}{x1 - x2} \\ = \frac{9 - 2}{5 - 2} \\ = \frac{7}{3} [/tex]

hope this helps

brainliest appreciated

good luck! have a nice day!

Answer:

Step-by-step explanation:

tan x-cot x

[tex]=\frac{sin~x}{cos~x} -\frac{cos~x}{sin ~x} \\=\frac{sin^2x-cos^2 x}{sin ~x~cos~x} \\=\frac{-2(cos^2x-sin^2x)}{2 sin ~x~cos~x} \\=\frac{-2 cos~2x}{sin~2x} \\=-2 cos ~2x~cosec~2x[/tex]

Answer:

Expected Payoff ⇒ $ 1.50 ; Type in 1.50

Step-by-step explanation:

Considering that 1 out of the 100 tickets will have a probability of winning a 150 dollar prize, take a proportionality into account;

[tex]100 - Number of Tickets,\\1 - Number of Tickets You Can Enter,\\\\1 / 100 - Probability of Winning,\\$ 150 - Money Won,\\\\Proportionality - 1 / 100 = x / 150, where x - " Expected Payoff "\\\\1 / 100 = x / 150,\\100 * x = 150,\\\\Conclusion ; x = 1.5 dollars[/tex]

Thus, Solution ; Expected Payoff ⇒ $ 1.50

Step-by-step explanation:

[tex]r = \sqrt{ \frac{ax - p}{q + bx} } \\ {r}^{2} = \frac{ax - p}{q + bx} [/tex]

r² (q + bx) = ax - p

qr² + bxr² = ax - p

qr² + p = ax - bxr²

qr² + p = x (a - br²)

[tex]x = \frac{q {r}^{2} + p}{a - b {r}^{2} } [/tex]

Answer:

[tex]\displaystyle x=\frac{-P-\math{R}^2Q}{\math{R}^2b-a}[/tex]

Step-by-step explanation:

[tex]R=\sqrt{\frac{ax-P}{Q+bx}}[/tex]

[tex]\mathrm{Square\:both\:sides}[/tex]

[tex]R^2=\left(\sqrt{\frac{ax-P}{Q+bx}}\right)^2[/tex]

[tex]R^2=\frac{ax-P}{Q+bx}[/tex]

[tex]\mathrm{Multiply\:both\:sides\:by\:}Q+bx[/tex]

[tex]\math{R}^2\left(Q+bx\right)=\frac{ax-P}{Q+bx}\left(Q+bx\right)[/tex]

[tex]\math{R}^2\left(Q+bx\right)=ax-P[/tex]

[tex]\math{R}^2Q+\math{R}^2bx=ax-P[/tex]

[tex]\mathrm{Subtract\:}\math{R}^2Q\mathrm{\:from\:both\:sides}[/tex]

[tex]\math{R}^2Q+\math{R}^2bx-\math{R}^2Q=ax-P-\math{R}^2Q[/tex]

[tex]\math{R}^2bx=ax-P-\math{R}^2Q[/tex]

[tex]\mathrm{Subtract\:}ax\mathrm{\:from\:both\:sides}[/tex]

[tex]\math{R}^2bx-ax=ax-P-\math{R}^2Q-ax[/tex]

[tex]\math{R}^2bx-ax=-P-\math{R}^2Q[/tex]

[tex]\mathrm{Factor}\:\math{R}^2bx-ax[/tex]

[tex]x\left(\math{R}^2b-a\right)=-P-\math{R}^2Q[/tex]

[tex]\mathrm{Divide\:both\:sides\:by\:}\math{R}^2b-a[/tex]

[tex]\frac{x\left(\math{R}^2b-a\right)}{\math{R}^2b-a}=-\frac{P}{\math{R}^2b-a}-\frac{\math{R}^2Q}{\math{R}^2b-a}[/tex]

[tex]x=\frac{-P-\math{R}^2Q}{\math{R}^2b-a}[/tex]