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3 years ago

Determine if the given ordered pair is a solution to the system of linear equations.

Mathematics

2 answers:

Molodets [167]3 years ago

7 0

For 1st equation: 2(7)+6 = 14+6 = 20 = R.H.S

For 2nd equation: 6(7)-5(6)= 42-30 = 12 = R.H.S.

So, your answer is "YES" for both system of equations.

Natalija [7]3 years ago

6 0

Above answer is correct. You can believe that

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the altitude of A (in feet) of a plane T minutes after liftoff is given by a=3400t + 600. How many minutes after liftoff is the

____ [38]

If our equation is $a = 3400t + 600$ , and we have a target altitude of $21,000$ , we just need to plug that in for $a$ .

$21,000 = 3,400t + 600$ . Now subtract $600$ from both sides.

$20,400 = 3,400t$ . now divide both sides by $3,400$ .

$6 = t$ .

We get that our answer is at 6 minutes we get an altitude of 21,000 feet!

6 0

3 years ago

Solve using the quadratic formula:

timurjin [86]

Answer:

$\boxed{x_1 = \frac{7 + \sqrt{5} }{2} ~~or~~x_2 = \frac{7 - \sqrt{5} }{2} }$

Step-by-step explanation:

$x^2 - 7x + 11 = 0$

$x^2 - 7x = -11$

$(x - \frac{7}{2} )^2 - \frac{49}{4} = -11$

$(x - \frac{7}{2} )^2 = -11 + \frac{49}{4}$

$(x - \frac{7}{2} )^2 = \frac{5}{4}$

$x - \frac{7}{2} = \pm \sqrt{\frac{5}{4}}$

$x = \frac{7 \pm \sqrt{5} }{2}$

3 0

3 years ago

Can I PLEASE get some help?

V125BC [204]

The 1st and 3rd choices are irrational. The other 2 are rational. Irrational numbers are numbers that you can't make a simple fraction out of.

7 0

3 years ago

Can anyone please help asap I keep getting those links bots

Zolol [24]

I don’t understand the question 14=7

5 0

3 years ago

Prove that.<br><br>lim Vx (Vx+ 1 - Vx) = 1/2 X>00

faltersainse [42]

Answer:

The idea is to transform the expression by multiplying $(\sqrt{x + 1} - \sqrt{x})$ with its conjugate, $(\sqrt{x + 1} + \sqrt{x})$ .

Step-by-step explanation:

For any real number $a$ and $b$ , $(a + b)\, (a - b) = a^{2} - b^{2}$ .

The factor $(\sqrt{x + 1} - \sqrt{x})$ is irrational. However, when multiplied with its square root conjugate $(\sqrt{x + 1} + \sqrt{x})$ , the product would become rational:

$\begin{aligned} & (\sqrt{x + 1} - \sqrt{x}) \, (\sqrt{x + 1} + \sqrt{x}) \\ &= (\sqrt{x + 1})^{2} -(\sqrt{x})^{2} \\ &= (x + 1) - (x) = 1\end{aligned}$ .

The idea is to multiply $\sqrt{x}\, (\sqrt{x + 1} - \sqrt{x})$ by $\displaystyle \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}$ so as to make it easier to take the limit.

Since $\displaystyle \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}} = 1$ , multiplying the expression by this fraction would not change the value of the original expression.

$\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \lim\limits_{x \to \infty} \left[\sqrt{x} \, (\sqrt{x + 1} - \sqrt{x})\cdot \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}\right] \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}\, ((x + 1) - x)}{\sqrt{x + 1} + \sqrt{x}} \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}}\end{aligned}$ .

The order of $x$ in both the numerator and the denominator are now both $(1/2)$ . Hence, dividing both the numerator and the denominator by $x^{(1/2)}$ (same as $\sqrt{x}$ ) would ensure that all but the constant terms would approach $0$ under this limit:

$\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \cdots\\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}} \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x} / \sqrt{x}}{(\sqrt{x + 1} / \sqrt{x}) + (\sqrt{x} / \sqrt{x})} \\ &= \lim\limits_{x \to \infty}\frac{1}{\sqrt{(x / x) + (1 / x)} + 1} \\ &= \lim\limits_{x \to \infty} \frac{1}{\sqrt{1 + (1/x)} + 1}\end{aligned}$ .

By continuity:

$\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \cdots\\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}} \\ &= \cdots \\ &= \lim\limits_{x \to \infty} \frac{1}{\sqrt{1 + (1/x)} + 1} \\ &= \frac{1}{\sqrt{1 + \lim\limits_{x \to \infty}(1/x)} + 1} \\ &= \frac{1}{1 + 1} \\ &= \frac{1}{2}\end{aligned}$ .

8 0

3 years ago

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