Select the correct answer from each drop down menu y=x^2+2x-1 y-3x=5

1 answer:

tatiyna3 years ago

4 0

The drop down menus are not provided. However, I will help get the solution of the system of equation which is the requirement of the question.

Answer:
(3,14) and (-2,-1) are the solutions of the system

Explanation:
The first equation given is:
y = x² + 2x - 1 .........> equation I
The second given equation is:
y - 3x = 5
This can be rewritten as:
y = 3x + 5 ..........> equation II

Equate equations I and II and solve for x as follows:
x² + 2x - 1 = 3x + 5
x² + 2x - 1 - 3x - 5 = 0
x² - x - 6 = 0
(x-3)(x+2) = 0

either x-3 = 0 .........> x = 3
or x+2 = 0 ...........> x = -2

Now, we get the values of y by substituting in equation II as follows:
At x = 3:
y = 3x + 5
y = 3(3) + 5
y = 14
The first solution is (3,14)

At x = -2:
y = 3x + 5
y = 3(-2) + 5
y = -1
The second solution is (-2,-1)

Hope this helps :)

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ax^2 + bx + c

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Make 'h' as the subject. Remember to rationalize the denominator while solving.

Sliva [168]

Hello there BeGuiLE !

To solve your question, we must first bring 'h' to one side of the equation. Let's bring all the terms with the variable 'h' to the left side of the equation. So,
$\large{\sf\sqrt{ 3 } h = 1.6+h}$
→ $\large{\sf\sqrt{3}h-h=1.6 }$

Now, let's combine the 2 terms containing the variable 'h'.
$\large{\sf\sqrt{3}h-h=1.6 }$
→ $\large{\sf\left(\sqrt{3}-1\right)h=1.6 }$

Now, we need to leave the variable 'h' alone in the left side of the equation & bring (√3 + 1) to the other side of the equation. This is done in order to make 'h' as the subject of the equation. So, we'll get it as,
$\large{\sf\left(\sqrt{3}-1\right)h=1.6 }$
→ $\large{\sf\:h=\frac{1.6}{\sqrt{3}-1} }$

Now, let's rationalize it. Rationalizing is a process where we move the root from the denominator of a fraction to the numerator for easier calculation. So,
$\large{\sf\:h=\frac{1.6}{\sqrt{3}-1} }$
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1 )(\sqrt{3}+1}) }$
Now, use the algebraic identity, (a + b)(a - b) = a² - b²
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{3-1} }$
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{2} }$
→ $\boxed{\large{\sf\:h=0.8(\sqrt{3}+1)}}$

Now, we've made 'h' as the subject of the equation. If you want to solve it more, then,
$\large{\sf\:h=0.8(\sqrt{3}+1)}$
Take √3 as 1.73 (approx. value up to 2 decimal places)
→ $\large{\sf\:h=0.8(1.73+1)}$
→ $\large{\sf\:h=0.8(2.73)}$
→ $\boxed{\boxed{\huge{\bf{\:h=2.184 \: (approx.)}}}}$

I hope this will help you.