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3 years ago

7

Sara is getting ready for work. She has black pants, blue pants, grey pants, a red shirt, a green shirt, a white shirt, a blue s

weater, and a grey sweater. If she chooses one pair of pants, one shirt, and one sweater, how many different outfits can she make?
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1 answer:

Nady [450]3 years ago

7 0

Answer:

3 pants

3 shirts

2 sweaters

just multiply 3x3x2= 18 combinations

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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:

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

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3 years ago

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If i got a phone with scratches on the back there must be light once’s on the front as well right? my phone has a few scratches

Gemiola [76]

I don’t think so the same thing happened to me so I think the screen protector was smaller than the actual phone that’s why there was scratches, next time you should look at the phone properly before buying it

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2 years ago

Solve: -1/2x +3 = -x + 7

motikmotik

Minus 3 both sides
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minus 2x both sides
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4 years ago

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Find the volume of a right circular cone that has a height of 12 cm and a base with a diameter of 18 cm.

algol [13]

Answer:1017.36 cm^3

Step-by-step explanation:

Diameter=18cm

Radius=r=18/2=9cm

Height=h=12cm

Volume of cone=1/3 x π x r^2 x h

Volume of cone=1/3 x 3.14 x 9^2 x 12

Volume of cone=1/3 3.14 x 9 x 9 x 12

Volume =(1x3.14x9x9x12) ➗ 3

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4 years ago

A student solved for the radius of a circle with a circumference of 12π Step 1: C = 2 pi r. Step 2: 12 pi = 2 pi r. Step 3: Star

Greeley [361]

Answer:

The student made a mistake in step 4

The length of the radius should be 6 cm

Step-by-step explanation:

Let us write each step to find the radius of a circle from its circumference

Step 1:

∵ C = 2πr

∵ C = 12π

Step 2:

Equate 12π by 2πr

∴ 12π = 2πr

Step 3:

Divide both sides by 2π

∵ $\frac{12\pi }{2\pi }=\frac{2\pi r}{2\pi }$

Step 4:

∴ 6 = r

∴ The radius of the circle is 6 cm

The student mad a mistake he divided the left hand side by 2π and the right hand side by 2 only he left π with r

Step 4:

∴ 6 = πr

So he divided both sides again by π

Step 5:

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Step 7:

∴ $\frac{6}{\pi }$ = r

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3 years ago

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