MATHEMATICS HIGH SCHOOL

Solve for d.
120 - 11d + 2d + 4d - 5 = 16

Answers

Answer 1

Answer:

Answer:

120 - 11d + 2d + 4d - 5 = 16

-5d = - 99

5d = 99

D = 19.8

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A circle has a radius of 11 inches and a central angle AOB that measures 45°. What is the area of sector AOB? Use 3.14 for pi and round your answer to the nearest tenth.a. 47.5 in2 b. 11.9 in2 c. 8.6 in2 d. 4.3 in2

Point m is the midpoint of line ab. if the coordinates of m are 2,8) and the coordinates of a are (10,12) what are the coordinates of b

Answers

midpoint = (2, 8)

coordinate a = (10. 12)

let coordinate b = (x, y)

Find x:

0.5(10 + x) = 2

10 + x = 4

x = 4 - 10

x = -6

Find y:

0.5(12 + y) = 8

12 + y = 16

y = 16 - 12

y = 4

Answer: Coordinate b = (-6, 4)

Find all real numbers X such that 6x-21<=-3 AND 14x+11>=-17

Answers

$6x - 21 \leq -3$
$14x + 11 \geq -17$

D: All real x.

$\text{From the first equation:}$
$6x \leq 18$
$x \leq 3$

New domain: x is less than or equal to 3.

$\text{From the second equation:}$
$14x \geq -28$
$x \geq -2$

Thus, there is a restriction:
$D: -2 \leq x \leq 3$

Please help me as soon as possible ...

Answers

Answer: see proof below

Step-by-step explanation:

Use the following identities

tan (A + B) = (tan A + tan B)/(1 - tan A · tan B) --> $(tanA+tanB)/(1-tanA\cdot tanB)$

tan 60° = √3

tan 120° = -√3

tan 3A = (3tanA - tan³A)/(1 - 3 tan²A) --> $(3tanA-tan^3A)/(1-3tan^2A)$

Proof LHS → RHS

Given: tan Ф + tan(60° + Ф) + tan(120° + Ф)

Sum Difference: tan Ф + (tan 60° + tanФ)/(1-tan60°·tanФ) + (tan 120° + tanФ)/(1-tan120°·tanФ)

(latex) $tan\theta+(tan60^o+tan\theta)/(1-tan60^o\cdot tan\theta)+(tan120^o+tan\theta)/(1-tan120^o\cdot tan\theta)$

Substitute: tan Ф + (√3 + tanФ)/(1-√3·tanФ) + (-√3 + tanФ)/(1+√3°·tanФ)

(latex) $tan\theta+(\sqrt3+tan\theta)/(1-\sqrt3 tan\theta)+(-\sqrt3+tan\theta)/(1+\sqrt3tan\theta)$

Common Denominator: [tan Ф(1-3tan²Ф)+8tanФ]\(1-3tan²Ф)

(latex) $(tan\theta(1-3tan^2\theta)+8\theta)/(1-3tan^2\theta)$

Distribute: (tan Ф - 3tan³Ф + 8Ф)\(1 - 3 tan²Ф)

(latex) $(tan\theta-3tan^3\theta+8\theta)/(1-3tan^2\theta)$

Simplify: (9Ф - 3tan³Ф)\(1 - 3 tan²Ф)

3(3Ф - tan³Ф)\(1 - 3 tan²Ф)

(latex) $(9\theta - 3tan^3\theta)/(1-3tan^2\theta)$

$(3(3\theta - tan^3\theta))/(1-3tan^2\theta)$

Triple Angle Identity: 3 tan 3Ф

3 tan 3Ф = 3 tan 3Ф $\checkmark$

Your question has been heard loud and clear.

tanθ+tan(60∘+θ)+tan(120∘+θ)

=tanθ+3–√+tanθ1−3–√tanθ+−3–√+tanθ1+3–√tanθ

[tanθ(1−3tan2θ)+(3–√+tanθ)(1+3–√tanθ)

=+(−3–√+tanθ)(1−3–√tanθ)1−3tan2θ

=9tanθ−3tan3θ1−3tan2θ=3tan3θ

Thank you.

A system of equations is shown below:y = 8x − 2
y = 9x − 7

What is the solution to the system of equations?

(−5, 38)
(−5, −38)
(5, 38)
(5, −38)

Answers

y = 8x - 2
y = 9x - 7

8x - 2 = 9x - 7
- 8x - 8x
-2 = x - 7
+ 7 + 7
5 = x

y = 8x -2
y = 8(5) - 2
y = 40 - 2
y = 38
(x, y) = (5, 38)

The answer is C.

Answer would be C (5, 38)

In a group of 55 people, 33 are coffee drinkers and 12 are tea drinkers. Assuming that no one in this group drinks both tea and coffee, what is the probability that a person picked at random from the group does not drink tea or coffee?

Answers

12+33=45. 45/55 people drink tea or coffee. This means if you picked someone randomly there would be a 45/55 chance that they drink tea or coffee. This means there is a 10/55 chance of someone who drinks neither. We can simplify this to 2/11.

Answer:

D. 2/11

Step-by-step explanation:

For all Plato users.

Does anybody know anything about simplifying rational expressions? 4x/11x