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san4es73 [151]
3 years ago
9

Help me plzz and It's continue of math

Mathematics
1 answer:
DedPeter [7]3 years ago
4 0

Answer: I only answered number 3

Step-by-step explanation: Jenni has a box of chocolates. The box contains 6 plain, 4 milk and 5 white chocolates.?

Jenni takes 2 chocolates at random from the box. Work out the probability that at least 1 of these chocolates will be milk chocolate.

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which is grater,the greatest whole number with 4 digits or the least whole number with 5 digits?explain.
Xelga [282]
The more digits you have the bigger the number, unless you have a long decimal, and in that case only is the number smaller in value.
5 0
3 years ago
-y= -2x +5 how do you make variable positive
Anastaziya [24]
Hej!

You would need to multiply both sides by -1

Resulting in...
y=2x-5
7 0
3 years ago
PLEASE ANSWER I GIVE BRAINLIEST write an equation of a line with m=6 that passes through (7,-5)
iren [92.7K]

Answer:

y = 6x - 47

or

6x - y - 47 = 0

Step-by-step explanation:

y -  y_{1} = m(x -  x_{1} ) \\  \\ plug \: x_{1} = 7 \\  y_{1} =  - 5 \\ m = 6 \: in \: the \: above \: equation \\ y - ( - 5) = 6(x - 7) \\  \\ y + 5 = 6x - 42 \\  \\ y = 6x - 42 - 5 \\  \\ y = 6x - 47 \\  \\ or \: 6x - y - 47 = 0

7 0
2 years ago
Which of the following values are solutions to the inequality 9 > - 3 - 4x
zvonat [6]

Answer:

the answer is 12

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Solving a trigonometric equation involving an angle multiplied by a constant
PIT_PIT [208]

In these questions, we need to follow the steps:

1 - solve for the trigonometric function

2 - Use the unit circle or a calculator to find which angles between 0 and 2π gives that results.

3 - Complete these angles with the complete round repetition, by adding

2k\pi,k\in\Z

4 - these solutions are equal to the part inside the trigonometric function, so equalize the part inside with the expression and solve for <em>x</em> to get the solutions.

1 - To solve, we just use algebraic operations:

\begin{gathered} \sqrt[]{3}\tan (3x)+1=0 \\ \sqrt[]{3}\tan (3x)=-1 \\ \tan (3x)=-\frac{1}{\sqrt[]{3}} \\ \tan (3x)=-\frac{\sqrt[]{3}}{3} \end{gathered}

2 - From the unit circle, we can see that we will have one solution from the 2nd quadrant and one from the 4th quadrant:

The value for the angle that give positive

+\frac{\sqrt[]{3}}{3}

is known to be 30°, which is the same as π/6, so by symmetry, we can see that the angles that have a tangent of

-\frac{\sqrt[]{3}}{3}

Are:

\begin{gathered} \theta_1=\pi-\frac{\pi}{6}=\frac{5\pi}{6} \\ \theta_2=2\pi-\frac{\pi}{6}=\frac{11\pi}{6} \end{gathered}

3 - to consider all the solutions, we need to consider the possibility of more turn around the unit circle, so:

\begin{gathered} \theta=\frac{5\pi}{6}+2k\pi,k\in\Z \\ or \\ \theta=\frac{11\pi}{6}+2k\pi,k\in\Z \end{gathered}

Since 5π/6 and 11π/6 are π radians apart, we can put them together into one expression:

\theta=\frac{5\pi}{6}+k\pi,k\in\Z

4 - Now, we need to solve for <em>x</em>, because these solutions are for all the interior of the tangent function, so:

\begin{gathered} 3x=\theta \\ 3x=\frac{5\pi}{6}+k\pi,k\in\Z \\ x=\frac{5\pi}{18}+\frac{k\pi}{3},k\in\Z \end{gathered}

So, the solutions are:

x=\frac{5\pi}{18}+\frac{k\pi}{3},k\in\Z

4 0
1 year ago
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