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3 years ago

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Find the solutions to x^2=27 A x= +/- 9 √3 B x= +/- 3 √9 C x= +/- 3 √3 D x= +/- 9 √2 FIRST TO ANSWER CORRECTLY GETS BRAINLIEST :

-)

2 answers:

Brilliant_brown [7]3 years ago

8 0

Answer:

The answer is C.

Step-by-step explanation:

You have to square both sides :

${x}^{2} = 27$

$x = ± \sqrt{27}$

Then, you have to simplify it :

$x = ± \sqrt{9 \times 3}$

$x = ±( \sqrt{9} \times \sqrt{3} )$

$x = ±( 3 \times \sqrt{3} )$

$x = ±3 \sqrt{3}$

GarryVolchara [31]3 years ago

7 0

Answer:

Option C) x = ± $3\sqrt{3}$

Step-by-step explanation:

=> $x^2 = 27$

Taking sqrt on both sides

=> x = ± $\sqrt{27}$

=> x = ± $\sqrt{3 * 3*3}$

=> x = ± $3\sqrt{3}$

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The formula for the circumference C of a circle with radius r is C = 2πr. Which of the following is a formula for π (pi) of the

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The best and most correct answer among the choices provided by your question is none of the above.
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olya-2409 [2.1K]

Answer:

(-4,9)

Step-by-step explanation:

To solve the system of equations, you want to be able to cancel out one of the variables. In this case, it'd be easiest to cancel out the x variables. To do this, you'll want to multiply everything in the first equation by 2 (2(x-5y=-49)=2x-10y=-98). Then, you can add the two equations together. 2x and -2x will cancel out, so you'll be left with -11y=-99. Next, solve for x by dividing both sides of the equation by -11, which will give you y=9. This is your y-coordinate! At this point, you're halfway to the answer as you just need your x-coordinate. It's not too difficult to find the x-coordinate, since you just substitute 9 into one of the equations. It doesn't matter which one you choose as you should get the same answer with both. I usually substitute the y-value into both equations, though, just to make sure I'm correct. Once you put the y-value into the equations, you should get x=-4 after solving it. :)

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3 years ago

If the original square had a side length of

irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$

$A2=(4)(x)=4x\ in^2$

$A3=(x)(7)=7x\ in^2$

$A4=(4)(7)=28\ in^2$

The total area of the second rectangle is the sum of the four areas

$A=A1+A2+A3+A4$

State the product of (x+4) and (x+7) as a trinomial

$(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28$

Part c) If the original square had a side length of x = 2 inches, then what is the area of the second rectangle?

we know that

The area of the second rectangle is equal to

$A=A1+A2+A3+A4$

For x=2 in

substitute the value of x in the area of each portion

$A1=(2)(2)=4\ in^2$

$A2=(4)(2)=8\ in^2$

$A3=(2)(7)=14\ in^2$

$A4=(4)(7)=28\ in^2$

$A=4+8+14+28$

$A=54\ in^2$

Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 in

We have that

The trinomial is

$A(x)=x^{2} +11x+28$

For x=2 in

substitute and solve for A(x)

$A(2)=2^{2} +11(2)+28$

$A(2)=4 +22+28$

$A(2)=54\ in^2$ ----> verified

therefore

The trinomial represent the total area of the second rectangle

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Step-by-step explanation:

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