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2 years ago

Evaluate √x^2 − y^2 when x^2 =33 and y^2 =15

Mathematics

1 answer:

Andre45 [30]2 years ago

3 0

Answer:

3√2

Step-by-step explanation:

In this question, we are asked to get the value of √ ( x²-y² ), while the value of x² and y² are given to be 33 and 15 respectively.

Hence, x² =33 and y² = 15.

Putting the values of x² and y² in the above expression we get,

√ ( x²-y² ) = √ ( 33-15) = √18 =3√2. (Answer)

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Andre45 [30]

Answer:

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3 years ago

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3 years ago

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You are training to compete in a 10-kilometer race, and you know the circular running trail at your park is one mile long. How m

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1 year ago

et f(x) = x + |2x − 3|. Find all solutions to the equation f(x) = 12. (Enter your answers as a comma-separated list. If an answe

Alinara [238K]

Answer:

1. {5,-9}

2.{4}

3.{9/2}

Step-by-step explanation:

The equation $f(x)=x+\lvert 2x-3 \rvert=12$ can be solved as follows:

$x+\lvert 2x-3 \rvert =12\,\,\,\text{(given equation)}$

$\lvert 2x-3 \rvert =12-x$

$2x-3=12-x\,\,\text{or}\,\,2x-3=-(12-x)$

$3x=15\,\,\text{or}\,\,\x=-9$

Then, the solutions of the equation are $x=5, x=-9$ .

Now, if $g(x)=5x-3+\lvert x+4\rvert$ , we can solve the equation $g(a)=4a+9$ as follows:

$5a-3+\lvert a+4 \rvert =4a+9$

$\lvert a+4\rvert =4a+9-5a+3$

$a+4 =-a+12\,\,\text{or}\,\,a+4=-(-a+12)$

$2a=8\,\,\text{or}\,\,4=12$

So, the unique solution of this last equation is $a=4$

The last equation is $h(x-1)=x-3$ , where $h(x)=\lvert x \rvert -2x+5.$ . To solve this equation we can proceed as follows:

$h(x-1)=x-3$

$\lvert x-1\rvert -2(x-1)+5=x-3$

$\lvert x-1 \rvert -2x+2+5=x-3$

$\lvert x-1\rvert =3x-10$

$x-1=3x-10\,\,\text{or}\,\,x-1=-(3x-10)$

$-2x=-9\,\,\text{or}\,\,4x=11$

$x=9/2\,\,\text{or}\,\,x=11/4$

But, x=11/4 is not a solution of this equation. So the unique solution is x=9/2.

8 0

2 years ago