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3 years ago

10 + (5 exponet 2 + 4 exponet 2)

Mathematics

2 answers:

Katen [24]3 years ago

6 0

Answer: 51

Step-by-step explanation:

10 + ( $5^{2}$ + $4^{2}$ )

10 + (25 + 16)

10 + (41)

10 + 41 = 51

hope this helps! if you have any questions, let me know!

Aleksandr [31]3 years ago

5 0

Answer:

Step-by-step explanation:

10 + (25 + 16)

10 + 41

=51

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A grocery store bought some mangoes at a rate of 5 for a dollar. They were separated into two stacks, one of which was sold at a

Alexus [3.1K]

Answer:

The ratio of the number of mangoes in the $3 stack to those in the $6 stack is 1 : 2

Step-by-step explanation:

Let the number of mangoes bought by the grocery store be n. Also let the number of mango sold for $3 in one stack be x and the number of mango sold for $6 in the second stack be y.

Therefore:

x + y = z (1)

Also, the mangoes was sold at break even price, that is the cost of the mango and the price it was sold for was the same. Therefore:

Cost of buying = Price it was sold for

The cost of the mango = 5z and the price it was sold for = 3x + 6y

3x + 6y = 5z (2)

Substituting z = x + y in equation 1

3x + 6y = 5(x + y)

3x + 6y = 5x + 5y

6y - 5y = 5x - 3x

y = 2x

x / y = 1/ 2 = 1 : 2

The ratio of the number of mangoes in the $3 stack to those in the $6 stack is 1 : 2

8 0

3 years ago

Can you Factor y^2+5y+6

fenix001 [56]

Hi there!

y² + 5y + 6

⇒ y² + 2y + 3y + 6

⇒ y (y + 2) + 3 (y + 2)

⇒ (y + 2) (y + 3)

~ Hope it helps!

6 0

4 years ago

Read 2 more answers

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:

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