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3 years ago

Solve using square roots 3x^2+25=73

1 answer:

4 0

Heya !

Given expression -

$3 {x}^{2} + 25 = 73$

Subtracting 25 both sides ,

$3 {x}^{2} + 25 - 25 = 73 - 25 \\ \\ 3 {x}^{2} = 48$

Dividing by 3 on both sides ,

$\frac{3 {x}^{2} }{3} = \frac{48}{3} \\ \\ {x}^{2} = 16$

Therefore ,
$x = + \: 4 \: \: \: \: or \: \: - 4 \: \: \: \: \: \: \: Ans.$

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Help me guys thx so much

navik [9.2K]

Answer:

Option D, $10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

Step-by-step explanation:

Step 1: Multiply

$(\sqrt{10x^4} -x\sqrt{5x^2} )*(2\sqrt{15x^4} + \sqrt{3x^3})\\ (\sqrt{10 * x^2 * x^2} -x\sqrt{5 * x^2} ) * (2\sqrt{15 * x^2 * x^2} +\sqrt{3 * x^2 * x})\\(x^2\sqrt{10} -x^2\sqrt{5} )*(2x^2\sqrt{15} +x\sqrt{3x}) \\\\$

$(x^2\sqrt{10}*2x^2\sqrt{15} )+(x^2\sqrt{10}*x\sqrt{3x} ) + (-x^2\sqrt{5} *2x^2\sqrt{15}) + (-x^2\sqrt{5} *x\sqrt{3x}$

$(2x^4\sqrt{150} ) + (x^3\sqrt{30x}) + (-2x^4\sqrt{75}) + (-x^3\sqrt{15x} )$

$2x^4\sqrt{5^2*6} + x^3\sqrt{30x} -2x^4\sqrt{5^2*3} -x^3\sqrt{15x}$

$10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

Answer: Option D, $10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

6 0

3 years ago

Read 2 more answers

6=p*36. **HINT** 17% IS NOT THE ANSWER AS TESTED!

grin007 [14]

Divide both sides by 36
6/36 = p
1/6 = p

3 0

3 years ago

There are 4 3/8 pounds of bricks in a bag. Each brick weighs 7/8 of a pound. How many bricks are in the bag?

goldfiish [28.3K]

Hello :) the answer to your question would be 5. There are 5 bricks in each bag. 35/8

7 0

1 year ago

Read 2 more answers

Sq is an angle bisector m <qst=2x and m <qsr=3x-10 what is <rst

defon

Answer:
∠RST = 40°

Explanation:
We are given that:
SQ bisects angle RST.
This means that:
∠QST = ∠QSR
2x = 3x - 10
10 = 3x - 2x
x = 10

Therefore:
∠QST = 2x = 2(10) = 20°
∠QSR = 3x - 10 = 3(10) - 10 = 30 - 10 = 20°
Note that both angles are equal.

Now, we can get ∠RST as follows:
∠RST = ∠QST + ∠QSR
∠RST = 20 + 20 = 40°

Hope this helps :)

6 0

3 years ago

prove that sin theta cos theta = cot theta is not a trigonometric identity by producing a counterexample

Zolol [24]

Answer:

To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.

Begin with the right hand side:

R.H.S = cot θ = $\frac{cos \ \theta}{sin \ \theta}$

L.H.S = sin θ cos θ

so, sin θ cos θ ≠ $\frac{cos \ \theta}{sin \ \theta}$

So, the equation is not a trigonometric identity.

=========================================================

Anther solution:

To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.

Assume θ with a value and substitute with it.

Let θ = 45°

So, L.H.S = sin θ cos θ = sin 45° cos 45° = (1/√2) * (1/√2) = 1/2

R.H.S = cot θ = cot 45 = 1

So, L.H.S ≠ R.H.S

So, sin θ cos θ = cot θ is not a trigonometric identity.

5 0

3 years ago