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3 years ago

14

Help me please I'm having trouble with this problem

1 answer:

sweet [91]3 years ago

7 0

The answer is (4,6) (4,3) (5,4)

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Find all solutions in the interval [0,2pi).<br> 2sin^2(x)=sin(x)

SIZIF [17.4K]

$2\sin^2 x =\sin x \\\\\implies 2 \sin^2 x - \sin x=0\\\\\implies \sin x(2 \sin x -1) =0\\\\\implies \sin x = 0~~\text{or}~~ 2 \sin x -1 =0\\\\\implies \sin x = 0~~ \text{or}~ ~ \sin x = \dfrac 12\\\\\text{For}~~ \sin x = 0\\\\x=n\pi \\\\\text{In the interval,}~~[0, 2\pi)\\\\x=0, \pi\\\\\text{For}~~ \sin x = \dfrac 12\\\\x=n\pi+(-1)^n \dfrac{\pi}6 \\\\\text{In the interval,}~~[0, 2\pi)\\\\x=\dfrac{\pi}6,~ \dfrac{5\pi}6\\\\\text{Hence,}~ x = 0,~\pi,~ \dfrac{\pi}6, ~\dfrac{5 \pi}6$

4 0

3 years ago

Read 2 more answers

What is the value for x?

expeople1 [14]

Answer:

X=8

Step-by-step explanation:

AB and AC are the same length so angles a and c are the same

180=a+b+c

180=69+b+69

b=42

42=5x+2

40=5x

8=x

4 0

4 years ago

The half-life of Carbon 14 (C-14) is 5230 years. Determine the decay-rate pa-rameterλfor C-14.

Lady bird [3.3K]

Answer:

λ = 1.3252 x 10⁻⁴

Step-by-step explanation:

Since we are already given the half-life, the decay expression can be simplified as:

$N(t) = N_0*e^{-\lambda t}\\\frac{N(half-life)}{N_0}=0.5$

For a half-life of t =5230 years:

$0.5 = e^{(-\lambda t)} \\ln(0.5) = -\lambda * 5230\\\lambda = 1.3252*10^{-4}$

The decay-rate parameter λ for C-14 is 1.3252 x 10⁻⁴

7 0

3 years ago

simplify 4 (5c ÷4d)-16 show me how you figure it out to put the numbers in this and I'll be asking two more questions for the ni

Darya [45]

$4(5c + 4d) - 16 \\ 20c + 16d - 16$
This is your final answer none of the terms are "like terms".

7 0

3 years ago

If y varies jointly as x and z and inversely as the square of w, and y = 3 when x = 3.2 = 10, and w = 2, then y - 4 when 7 - 4.2

kiruha [24]

Answer:

Following the equation in the attacked file you can calculate any of the quantity that is given

Step-by-step explanation:

3 0

4 years ago