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4 years ago

Can someone help me with these ?

Mathematics

1 answer:

abruzzese [7]4 years ago

5 0

You would plug in the probelm would be 3(30+8) = 2×30 -6

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Express the following ratios in their simplest form 24:15:9

11Alexandr11 [23.1K]

8:5:3 this is the simplest ratio

7 0

3 years ago

Tyrell wants to buy a pair of shoes with an original price of $75. He has a coupon for 25% off. What is the price of the shoes o

kirza4 [7]

$56.25, since 25% off can also be written as ÷4, since 25 is a quarter. 75÷4=18.75
75-18.75=56.25. Therefore, the answer is $56.25

3 0

3 years ago

Prove that. 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

Read 2 more answers

a kayaker travels x miles per hour downstream for 3 hours. on the 4-hour return trip, the kayaker travels 1 mile per hour slower

o-na [289]

3x is the miles from downstream
4 (x - 1) is the return trip

Solution:
3x = 4 (x - 1)
3x =4x - 4
-x = -4
x = 4

The outward journey 3x = 12 miles
Return journey is same length
Therefore the distance travel is 24 miles

7 0

3 years ago

Read 2 more answers

By using a Diophantine equation answer the following question. No credit will be given unless you show all your work. ll . A bou

zalisa [80]

Answer:

20 cardigans
30 pullovers

Step-by-step explanation:

Let c and p represent the numbers of cardigans and pullovers sold, respectively. Then the revenue equation is ...

31c +28p = 1460

Solving for p, we have ...

p = (1460 -31c)/28 = 52 -c +(4 -3c)/28

Now, define ...

a = (4 -3c)/28

and solve for c:

(4-28a)/3 = c = 1 -9a +(1 -a)/3

At this point, we can define ...

b = (1 -a)/3

Working backward, we can find c and p.

a = 1 -3b

c = 1 -9a +b = 1 -9(1 -3b) +b = -8 +28b

p = 52 -c +a = 52 -(-8 +28b) +(1 -3b) = 61 -31b

That is, for some integer b, solutions to the equation will be ...

c = 28b -8

p = 61 -31b

The only value of b that gives positive solutions for both c and p is b=1. For that value of b, we have ...

c = 20, p = 30

The boutique owner sold 20 cardigans and 30 pullovers.

_____

Comment on the solution

This is a variation of the Extended Euclidean Algorithm. Here, we have defined intermediate variables (a and b) that might not show up in a tableau solution form.

6 0

4 years ago