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8090 [49]
3 years ago
15

B can you help me ...​

Mathematics
1 answer:
Gemiola [76]3 years ago
8 0
Answer: a is an element of set A. But p is not an element of set A.

Explanation:
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6,484 ÷ 24 what is the quotient?
Mama L [17]
270.1666666666667     this is the anwser
                  
3 0
3 years ago
Read 2 more answers
What is the output of the function when the input is 0
bearhunter [10]

Answer:

y = -1

Step-by-step explanation:

Any ordinate pair (x, y) represents the input output values of a function to be graphed.

For any input value of x, there will be an output value (y).

If input value for the graph attached is x = 0,

Output value will be represented by the y-value along y-axis as, y = -1

Therefore, y = -1 will be the answer.

4 0
3 years ago
The scale on map 1inch:3miles The movies is 5 inches on the map from the house. How far is the movies actually from the house?
nignag [31]
15 miles cause you do 5 times 3 and you get 15 miles
7 0
3 years ago
If tan a = 3/4 find the value of sin 3a
Effectus [21]
T=-1

sinA=sin(π/2-3A), A=2nπ+π/2-3A, 4A=2nπ+π/2, A=nπ/2+π/8 where n is an integer.

Also, π-A=2nπ+π/2-3A, 2A=2nπ-π/2, A=nπ-π/4.

The hard way:

cos3A=cos(2A+A)=cos(2A)cosA-sin(2A)sinA.

Let s=sinA and c=cosA, then s²+c²=1.

cos3A=(2c²-1)c-2c(1-c²)=c(4c²-3).

s=c(4c²-3) is the original equation.

Let t=tanA=s/c, then c²=1/(1+t²).

t=4c²-3=4/(1+t²)-3=(4-3-3t²)/(1+t²)=(1-3t²)/(1+t²).

So t+t³=1-3t², t³+3t²+t-1=0=(t+1)(t²+2t-1).

So t=-1 is a solution.

t²+2t-1=0 is a solution, t²+2t+1-1-1=0=(t+1)²-2, so t=-1+√2 and t=-1-√2 are solutions.

Therefore tanA=-1, -1+√2, -1-√2 are the three solutions from which:

A=-π/4, π/8, -3π/8 radians and these values +2πn where n is an integer.

Replacing π by 180° converts the solutions to degrees.

3 0
3 years ago
Are the equation linear or nonlinear?​
I am Lyosha [343]

Answer:

<em>Results below</em>

Step-by-step explanation:

<u>Equation of the line</u>

A straight line can be written in the form:

y = ax + b

Where a and b are constants and x is the independent variable.

The essential condition for an equation to be linear is that the x must be powered to the exponent 1, which is usually not written.

From the equations presented in the table:

y = x^1+2 is linear because the exponent of the x is 1

y = 5(x+2) = 5x + 10 is linear with a=5 and b=10.

y = x is linear with a=1 and b=0

y = x^2+1 is not linear because the exponent of x is 2

y = x(x) = x^2 is not linear because the exponent of x is 2

The table below summarizes the results

8 0
2 years ago
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