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3 years ago

Can you help me with these questions please?

Mathematics

1 answer:

Marianna [84]3 years ago

5 0

The answers are following

sol1 y = x-1

sol2 y= x+ 7

sol4 y = x+4

Step-by-step explanation:

Solution 1

Here we are going put the point in the given equation ( given in the options)

If the each point satisfies any equation, that will be the relationship between x and y.

Given points are (-3,-4) , (2,1) , (5,4)

Let us take the option C as y = x -1

-4 = -3-1

1 = 2-1

4=5-1

Putting each value in the given equation we get LHS and RHS are same for all points. so, C is the answer.

Similarly, option B is the answer of Solution 2

Each point satisfies the equation y = x +7

Solution 4

y = x +4

You are requested to check the options of Q3 once.

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Which of the following is not equal to sin⁡(-230°)?

xenn [34]

From trigonometry we know that:

if $sin(\theta)=sin(\alpha)$

then, $\theta=n\pi+(-1)^n\alpha$ (where $n$ is an integer)

This can be rewritten in degrees as:

$\theta=n(180^{\circ})+(-1)^n\alpha$ .............(Equation 1)

Now, in our case, $\alpha=-230^{\circ}$

Therefore, (Equation 1) can be written as:

$\theta=n(180^{\circ})+(-1)^n(-230^{\circ})$ ..........(Equation 2)

Now, to find the correct options all that we have to do is replace n by relevant integers and find the values of $\theta$ that match.

For n=2, (Equation 2) gives us: $\theta=2\times 180^{\circ}+(-1)^2(-230^{\circ})=360^{\circ}-230^{\circ}=130^{\circ}$ .

Thus, $sin(230^{\circ})=sin(130^{\circ})$

Now, we know that: $-sin(-50^{\circ})=sin(50^{\circ})$

Let n=-1, then:

$\theta=(-1)\times 180^{\circ}+(-1)^{-1}(-230^{\circ})=-180^{\circ}+230^{\circ}=50^{\circ}$

Thus, $sin(-230^{\circ})=-sin(-50^{\circ})$

Likewise, $sin(-230^{\circ})=sin(50^{\circ})$

Only the last option $sin(-50^{\circ})$ will never match $sin(-230^{\circ})$ because no integral value of $n$ will ever give $\theta=-50^{\circ}$

Thus the last option is the correct option.

3 0

3 years ago

Read 2 more answers

Five distinct numbers are randomly distributed to players numbers 1 through 5. Whenever two playerscompare their numbers, the on

ad-work [718]

Answer:

Observe that the first and the second player have equal probability to get any of number. Using the principle of that symmetry, we have that

$P(X=0)=\frac{1}{2}$

Event X = 1 means that the first player has got greater number than the second player, but not than the third player. So, choose any three numbers out of five of them and say that the minimal number out of these three goes to the second player, mean number to the first one and the largest to the third one. Permute remaining two numbers on remaining two people. Hence

$P(X=1)=\frac{1}{6}$

Event X = 2 means that the first player has got greater number than the second and the third player, but not than the fourth player. So, choose any four numbers out of five of them and say that the minimal number and the next minimal out of these four go to the second and the third player (and permute them), third number to the first one and the largest to the fourth player. Give remaining number to the last person. Hence

$P(X=2)=\frac{1}{12}$

Event X = 3 means that the first player has got greater number than the second, the third, and the forth player, but not than the fifth player. So, permute these five numbers as follows: give the highest to the last person, the second highest to the first, and permute remaining numbers on the remaining people. Hence

$P(X=3)=\frac{1}{20}$