Solve: -2c = -18. 1)c= -92)c= -363)c= 94)c= 36

5.75 , 7.75 , 9.75 , 11.75 is a Non linear sequence ? is it true or false

Tomtit [17]

Answer:

false because each go up by two and I didnt even look it up

5 0

3 years ago

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This is easy i don’t know it tho

Ede4ka [16]

Answer:

5 in total

Step-by-step explanation:

7 0

4 years ago

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Which shows the correct substitution of the values a, b, and c from the equation 1 = -2x + 3x2 + 1 into the quadratic

lidiya [134]

Answer: A

Step-by-step explanation:

The first thing we need to do is to make sure our equation is in standard form. The given equation is not in standard form.

3x²-2x=0

Now that the equation is in standard form, we can find our A, B, C to see which quadratic equation is correct.

A=3

B=-2

C=0

We can automatically eliminate D because the first value is -3 when it is supposed to be -(-2). Also, the denominator is supposed to be 2a. We know that A=3. The denominator should be 2(3), not 2(-2).

Next, we can eliminate B and C. For the 4ac, we know it has to be 4(3)(0) because A=3 and C=0. B and C have given 4(3)(2) and 4(3)(1), respectively. This does not match A=3 and C=0.

Therefore, A is the correct answer.

4 0

4 years ago

3x + 5 =11 use the subtraction property of equality

notka56 [123]

2 subtract 5 from both sides, cancel out 5 in left. 3x=6 divide 3 from both sides so x=2

5 0

3 years ago

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Q 2 PLEASE HELP ME FIGURE THIS OUT

suter [353]

Answer: IV, positive, $\frac{\pi} {6}$ , - sec $\frac{\pi} {6}$ , $\frac{2\sqrt{3}}{3}$

Step-by-step explanation:

a) Look at the Unit Circle to see that $\frac{11\pi} {6}$ = 330°, which is located in Quadrant IV.

b) The coordinate (cos θ, sin θ) for $\frac{11\pi} {6}$ is: $(\frac{\sqrt{3}} {2},\frac{-1}{2})$

sec = $\frac{1}{cos}$ = $\frac{2}{\sqrt{3}}$ which is positive

c) Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the given angle $\frac{11\pi} {6}$ from 2π: $\frac{12\pi} {6}$ - $\frac{11\pi} {6}$ = $\frac{\pi} {6}$

d) the reference angle is below the x-axis so the given angle is equal to the negative of the reference angle: - sec $\frac{\pi} {6}$ .

e) sec $\frac{11\pi} {6}$ = $\frac{2}{\sqrt{3}}$ = $\frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}$ = $\frac{2\sqrt{3}}{3}$

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Answer: $\frac{18\pi}{11}$ , IV, $\frac{4\pi} {11}$

Step-by-step explanation:

2π is one rotation. 2π = $\frac{22\pi}{11}$

$\frac{-26\pi}{11}$ + $\frac{22\pi}{11}$ = $\frac{-4\pi}{11}$

$\frac{-4\pi}{11}$ + $\frac{22\pi}{11}$ = $\frac{18\pi}{11}$

Convert the radians into degrees to see which Quadrant it is in by setting up the proportion and cross multiplying:

$\frac{\pi}{180}$ = $\frac{18\pi}{11x}$

π(11x) = (180)18π

x = $\frac{180(18\pi}{11\pi}$

x = 295° which lies in Quadrant IV

Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the angle of least nonegative value $\frac{18\pi} {11}$ from 2π: $\frac{22\pi} {11}$ - $\frac{18\pi} {11}$ = $\frac{4\pi} {11}$

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Answer: $\frac{5\pi}{3}$ , IV, $\frac{4\pi} {11}$ , $\frac{\pi} {3}$

Step-by-step explanation:

2π is one rotation. 2π = $\frac{6\pi}{3}$

$\frac{-13\pi}{3}$ + $\frac{6\pi}{3}$ = $\frac{-7\pi}{3}$

$\frac{-7\pi}{3}$ + $\frac{6\pi}{3}$ = $\frac{-\pi}{3}$

$\frac{-\pi}{3}$ + $\frac{6\pi}{3}$ = $\frac{5\pi}{3}$

This is on the Unit Circle at 300°, which is located in Quadrant IV

Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the angle of least nonegative value $\frac{5\pi} {3}$ from 2π: $\frac{6\pi} {3}$ - $\frac{5\pi} {3}$ = $\frac{\pi} {3}$

7 0

4 years ago