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Jobisdone [24]
3 years ago
12

Find the exact value of each of the following. In each case, show your work and explain the steps you take to find the value.

Mathematics
1 answer:
mote1985 [20]3 years ago
4 0

Answer: a) \frac{1}{2}

b) 1

c) 2

Step-by-step explanation:

(a) sin 17π/6  

It is known that the value of sin x repeat after an interval of 2\pi\ or\ 360^{\circ}

∴ \sin\frac{17\pi}{6}=\sin2\frac{5\pi}{6}=\sin(2\pi+\frac{5}{6}\pi)=\sin{\frac{5}{6}\pi}=\sin(\pi-\frac{\pi}{6})=\sin(\frac{\pi}{6})=\frac{1}{2}

[Since the value of sin x is positive in 2nd quadrant]

(b) tan 13π/4  

It is known that the value of sin x repeat after an interval of \pi\ or\ 180^{\circ}

∴ \tan\frac{13\pi}{4}=\tan(3\pi+\frac{\pi}{4})=\tan{\frac{\pi}{4}}=1

(c) sec 11π/3

\text{Since, }\sec(x)=\frac{1}{\cos x}

\cos(\frac{11\pi}{3})=cos(\frac{5\pi}{3}+2\pi)=\cos(\frac{5\pi}{3})=\cos(6\pi-\frac{\pi}{3})=\cos(\frac{\pi}{3})=\frac{1}{2}\\\Rightarrow\sec(\frac{11\pi}{3})=\frac{1}{\cos(\frac{11\pi}{3})}=2

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Write as a product 27a^3−(a−b)^3
jarptica [38.1K]

Answer:

  (2a +b)·(13a^2 -5ab +b^2)

Step-by-step explanation:

The factorization of the difference of cubes is a standard form:

  (p -q)^3 = (p -q)(p^2 +pq +q^2)

Here, you have ...

  • p = 3a
  • q = (a-b)

so the factorization is ...

  (3a -(a -b))·((3a)^2 +(3a)(a -b) +(a -b)^2) . . . . substitute for p and q

  = (2a +b)·(9a^2 +3a^2 -3ab +a^2 -2ab +b^2) . . . . simplify a bit

  = (2a +b)·(13a^2 -5ab +b^2) . . . . . . collect terms

6 0
3 years ago
If the perpendicular bisector of one side of a triangle goes through the opposite vertex, then is the triangle isosceles?
Ne4ueva [31]

Answer:

True

Step-by-step explanation:

The perpendicular bisector of the opposite side to the vertex bisects the angle at the vertex into two equal parts and also bisects the triangle into two equal parts.

Let A be the angle at the vertex, then assume that the angle is an isosceles triangle with base angles B.

We need to show that A = 180 - 2B for an isoceles triangle

The perpendicular bisector bisects A into two so the new angle in the vertex one half of the bisected triangle is A/2.

Since this half triangle is a right-angled triangle, the third angle in it is 90.

So, A/2 + B +  90 = 180 (Sum of angles in a triangle)

subtracting 90 from both sides, we have

A/2 + B + 90 - 90 = 180 - 90

A/2 + B = 90

subtracting B from both sides, we have

A/2 + B = 90

A/2 = 90 - B

multiplying through by 2, we have

A = 2(90 - B)

A = 180 - 2B

Since A = 180 - 2B, then our triangle is an isosceles triangle.

7 0
3 years ago
In B E C, F is the centroid, and AC = 12. Find A F and F C
klemol [59]

bearing in mind that the centroid in a triangle cuts each of the three medians in a 2:1 ratio.

since we know that A C = 12, let's split it in a 2:1 ratio then, cleary from the picture the larger is F C, so F C : A F is on a 2:1 ratio.

\bf A C=A F+F C\qquad \qquad \cfrac{F C}{A F}=\cfrac{2}{1}\qquad \qquad \cfrac{2\cdot \frac{12}{2+1}}{1\cdot \frac{12}{2+1}}\implies \cfrac{2\cdot 4}{1\cdot 4}\implies \cfrac{8}{4}

4 0
3 years ago
Status
blsea [12.9K]

Answer:

y = -2x - 1

Step-by-step explanation:

y + 3 = -2 ( x - 1 )

y + 3 = -2x + 2

y = -2x - 1

5 0
2 years ago
Can someone please help me on number 16-ABC
melomori [17]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

-2x < 10

-6 < -2x

<u>Part a) Is x = 0 a solution to both inequalities</u>

FOR  -2x < 10

substituting x = 0 in -2x < 10

-2x < 10

-3(0) < 10

0 < 10

TRUE!

Thus, x = 0 satisfies the inequality -2x < 10.

∴ x = 0 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 0 in -6 < -2x

-6 < -2x

-6 < -2(0)

-6 < 0

TRUE!

Thus, x = 0 satisfies the inequality -6 < -2x

∴ x = 0 is the solution to the inequality -6 < -2x

Conclusion:

x = 0 is a solution to both inequalites.

<u>Part b) Is x = 4 a solution to both inequalities</u>

FOR  -2x < 10

substituting x = 4 in -2x < 10

-2x < 10

-3(4) < 10

-12 < 10

TRUE!

Thus, x = 4 satisfies the inequality -2x < 10.

∴ x = 4 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 4 in -6 < -2x

-6 < -2x

-6 < -2(4)

-6 < -8

FALSE!

Thus, x = 4 does not satisfiy the inequality -6 < -2x

∴ x = 4 is the NOT a solution to the inequality -6 < -2x.

Conclusion:

x = 4 is NOT a solution to both inequalites.

Part c) Find another value of x that is a solution to both inequalities.

<u>solving -2x < 10</u>

-2x\:

Multiply both sides by -1 (reverses the inequality)

\left(-2x\right)\left(-1\right)>10\left(-1\right)

Simplify

2x>-10

Divide both sides by 2

\frac{2x}{2}>\frac{-10}{2}

x>-5

-2x-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}

<u>solving -6 < -2x</u>

-6 < -2x

switch sides

-2x>-6

Multiply both sides by -1 (reverses the inequality)

\left(-2x\right)\left(-1\right)

Simplify

2x

Divide both sides by 2

\frac{2x}{2}

x

-6

Thus, the two intervals:

\left(-\infty \:,\:3\right)

\left(-5,\:\infty \:\right)

The intersection of these two intervals would be the solution to both inequalities.

\left(-\infty \:,\:3\right)  and \left(-5,\:\infty \:\right)

As x = 1 is included in both intervals.

so x = 1 would be another solution common to both inequalities.

<h3>SUBSTITUTING x = 1</h3>

FOR  -2x < 10

substituting x = 1 in -2x < 10

-2x < 10

-3(1) < 10

-3 < 10

TRUE!

Thus, x = 1 satisfies the inequality -2x < 10.

∴ x = 1 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 1 in -6 < -2x

-6 < -2x

-6 < -2(1)

-6 < -2

TRUE!

Thus, x = 1 satisfies the inequality -6 < -2x

∴ x = 1 is the solution to the inequality -6 < -2x.

Conclusion:

x = 1 is a solution common to both inequalites.

7 0
3 years ago
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