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3 years ago

Please solve this!!!!!

Mathematics

1 answer:

vfiekz [6]3 years ago

4 0

Answer:

(1, 2 )

Step-by-step explanation:

Given the 2 equations

3x + y = 5 → (1)

5x - 4y = - 3 → (2)

Rearrange (1) expressing y in terms of x by subtracting 3x from both sides

y = 5 - 3x → (3)

Substitute y = 5 - 3x into (2)

5x - 4(5 - 3x) = - 3 ← distribute and simplify left side

5x - 20 + 12x = - 3

17x - 20 = - 3 ( add 20 to both sides )

17x = 17 ( divide both sides by 17 )

x = 1

Substitute x = 1 into (3) for corresponding value of y

y = 5 - 3(1) = 5 - 3 = 2

Solution is (1, 2 )

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3 years ago

Solve for x. 2=4x-6

77julia77 [94]

Answer:

x = 2

Step-by-step explanation:

Rewrite the equation as 4x - 6 = 2

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Divide each term in 4x = 8 by 4.

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3 years ago

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3 years ago

PLEASE HELP!

sesenic [268]

Answer:

$36\sqrt{3}$

Step-by-step explanation:

Hi!

Since an equilateral triangle means that every side is equal, our triangle will have 12 on all sides.

To find the height of an equilateral triangle we use $\frac{a\sqrt{3}}{2}$ .

$\frac{12\sqrt{3} }{2} = 6\sqrt{3}$ .

So the height is $6\sqrt{3}$ .

Now we have to solve 12 * $6\sqrt{3}$ ÷ 2.

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Thus, the area of the triangle is $\boxed{36\sqrt{3}}$ .

Hope this helps!

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3 years ago

What is the following sum? Assume x20 and 20.<br> √√x²y³ +2√√x³y² + xy √√y

max2010maxim [7]

The value of the expression $\sqrt{x^2y^3} + 2\sqrt{x^3y^4} + xy\sqrt y$ is $2xy\sqrt{y} + 2xy^2\sqrt{x}$

<h3>How to evaluate the sum?</h3>

The attachment represents the proper format of the question

The summation expression is given as:

$\sqrt{x^2y^3} + 2\sqrt{x^3y^4} + xy\sqrt y$

Expand the radicands

$\sqrt{x^2y^2 * y} + 2\sqrt{x^2y^4* x} + xy\sqrt y$

Evaluate the square roots

$xy\sqrt{y} + 2xy^2\sqrt{x} + xy\sqrt y$

Add the like terms

$2xy\sqrt{y} + 2xy^2\sqrt{x}$

Hence, the value of the expression $\sqrt{x^2y^3} + 2\sqrt{x^3y^4} + xy\sqrt y$ is $2xy\sqrt{y} + 2xy^2\sqrt{x}$