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denis-greek [22]

3 years ago

8

Given circle P, which of the following options are major arcs?

1 answer:

MakcuM [25]3 years ago

3 0

If your asking to select all that apply,

The formula is arc length = θ ( r ) {\displaystyle {\text{arc length}}=\theta (r)} , where equals the measurement of the arc's central angle in radians, and r {\displaystyle r} equals the length of the circle's radius. Plug the length of the circle's radius into the formula.

ABC,BAD,DCB

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Use the grouping method to factor this polynomial completely.<br> 3x3 + 12x2 + 2x + 8

dezoksy [38]

Answer:

(3x^2 + 2)(x + 4).

Step-by-step explanation:

3x3 + 12x2 + 2x + 8

3x^2(x + 4) + 2(x + 4) The x + 4 is common to the 2 groups so we have:

(3x^2 + 2)(x + 4).

3 0

3 years ago

What is the solution to the following equation?

lions [1.4K]

Answer:

D. x = 7

Step-by-step explanation:

NOTE : <u><em>there should be an equal sign somewhere in the given expression.</em></u>

suppose the equation is the following:

3(x-4)-5 = x-3

………………………………………………………

3(x - 4) - 5 = x - 3

⇔ 3x - 12 - 5 = x - 3

⇔ 3x - 17 = x - 3

⇔ 3x - 17 + 17= x - 3 + 17

⇔ 3x = x + 14

⇔ 2x = 14

⇔ x = 14/2

⇔ x = 7

8 0

2 years ago

Estimate 25.65 + 13.23 + 6.35 by rounding each number to the nearest tenth.

sergij07 [2.7K]

Answer:

b. 25.7 + 13.2 + 6.4 = 45.3

Step-by-step explanation:

25.65 ≈ 25.7, 13.23 ≈ 13.2, 6.35 ≈ 6.4

25.65 + 13.23 + 6.35 =
25.7 + 13.2 + 6.4 =
45.3

Correct choice is b.

5 0

3 years ago

Which of the following is a solution to 2cos2x − cos x − 1 = 0?

Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, $x = 0^{\circ}$

Step-by-step explanation:

Given that : $2\cos^2x -\cos x -1 =0$

Let $\cos x =y$

then our equation become;

$2y^2-y-1= 0$ .....[1]

A quadratic equation is of the form:

$ax^2+bx+c =0$ .....[2] where a, b and c are coefficient and the solution is given by;

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

$y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}$

Simplify:

$y = \frac{ 1 \pm \sqrt{1+8}}{4}$

or

$y = \frac{ 1 \pm \sqrt{9}}{4}$

$y = \frac{ 1 \pm 3}{4}$

or

$y = \frac{1+3}{4}$ and $y = \frac{1 -3}{4}$

Simplify:

y = 1 and $y = -\frac{1}{2}$

Substitute y = cos x we have;

$\cos x = 1$

⇒ $x = 0^{\circ}$

and

$\cos x = -\frac{1}{2}$

⇒ $x = 120^{\circ} \text{and} x = 240^{\circ}$

The solution set: $\{0^{\circ}, 120^{\circ} , 240^{\circ}\}$

Therefore, the solution for the given equation $2\cos^2x -\cos x -1 =0$ is, $0^{\circ}$

8 0

3 years ago

Read 2 more answers

Find the solution(s) to the system <br>y = x2-4 <br>y = -2x – 5 <br><br>

katen-ka-za [31]

Answer:

=(-1,-5)

Step-by-step explanation:

y=X1=4

y=-2x-5-2

4 0

2 years ago