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3 years ago

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What is the recursive formula for each arithmetic sequence. -5,-8.5,-12,-15.5,-19

1 answer:

dusya [7]3 years ago

5 0

The answer is in the image

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Slope of the line (5,4) and (8,2)

GenaCL600 [577]

2/-3 by using the formula y2-y1/x2-x1

5 0

3 years ago

How to factor 3x^2-10x-8?

Soloha48 [4]

<em> $answer \\(x - 4)(3x + 2) \\ solution \\ {3x}^{2} - 10x - 8 \\ = {3x}^{2} - (12 - 2)x - 8 \\ = {3x}^{2} - 12x + 2x - 8 \\ = 3x(x - 4) + 2(x - 4) \\ = (x - 4)(3x + 2) \\ hope \: it \: helps$ </em>

4 0

4 years ago

Read 2 more answers

Ruben adds some water to a glass of juice. The ratio of water to juice in his glass in now 4:6. Which fraction below represents

zimovet [89]

Answer:

C

Step-by-step explanation:

since we're being told to convert the ratio to fraction then it is C. note that 4 ratio is also the same as dividing 4 by 6

6 0

3 years ago

Save me the headache

maxonik [38]

$(9\sin2x+9\cos2x)^2=81$

Taking the square root of both sides gives two possible cases,

$9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1$

or

$9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1$

Recall that

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

If $\alpha=2x$ and $\beta=\dfrac\pi4$ , we have

$\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}$

so in the equations above, we can write

$\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1$

Then in the first case,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi$

(where $n$ is any integer)

$\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi$

$\implies x=n\pi\text{ or }\dfrac\pi4+n\pi$

and in the second,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi$

$\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi$

$\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi$

Then the solutions that fall in the interval $[0,2\pi)$ are

$x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4$

5 0

3 years ago

Read 2 more answers

Suppose the discrete random variable X has the probability distribution below:

Talja [164]

(a) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.11 + 0.52 + 0.19 = 0.82

(b) P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.52 + 0.19 + 0.12 + 0.06 = 0.89

(c) µ = 0×0.11 + 1×0.52 + 2×0.19 + 3×0.12 + 4×0.06 = 1.5

(d) σ² = (0²×0.11 + 1²×0.52 + 2²×0.19 + 3²×0.12 + 4²×0.06) - µ² = 1.07

σ = √(σ²) ≈ 1.03

4 0

3 years ago