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2 years ago

Please need help thank

Mathematics

2 answers:

DIA [1.3K]2 years ago

8 0

Answer: 3 apples left

65 devided by 5 13

13-10

Step-by-step explanation:

stiv31 [10]2 years ago

3 0

Answer:

3 apples

Step-by-step explanation:

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Round .1325 to nearest thousandths

m_a_m_a [10]

So,

If the number in the ten-thousandth's place is greater than or equal to 5, then round up. If not, round down.

0.1325: Note the 5. We round up.

0.1325 --> 0.133

4 0

2 years ago

Read 2 more answers

2 questions when can you combine like terms? when can you NOT combine like terms?

Rainbow [258]

*just a guess*

you can combine like terms when they are using the same variables or constants. ex: you can combine 34x and 76x because of the variable

6 0

2 years ago

Use the identity (x2+y2)2=(x2−y2)2+(2xy)2 to determine the sum of the squares of two numbers if the difference of the squares of

fomenos

Answer:

The sum of the squares of two numbers whose difference of the squares of the numbers is 5 and the product of the numbers is 6 is 169

Step-by-step explanation:

Given : the difference of the squares of the numbers is 5 and the product of the numbers is 6.

We have to find the sum of the squares of two numbers whose difference and product is given using given identity,

$(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2$

Since, given the difference of the squares of the numbers is 5 that is $(x^2-y^2)^2=5$

And the product of the numbers is 6 that is $xy=6$

Using identity, we have,

$(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2$

Substitute, we have,

$(x^2+y^2)^2=(5)^2+(2(6))^2$

Simplify, we have,

$(x^2+y^2)^2=25+144$

$(x^2+y^2)^2=169$

Thus, the sum of the squares of two numbers whose difference of the squares of the numbers is 5 and the product of the numbers is 6 is 169

5 0

2 years ago

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The table shows the relationship between the number that fit into given number of boxes from a bakery . What is the missing valu

ki77a [65]

Answer:

The missing value should be 48 because between the two numbers of each group. They are multplied by 6

Step-by-step explanation:

Hope this Helped!

7 0

2 years ago

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What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$

∴ we have evaluated the given limit.

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Learn more about limits: brainly.com/question/27807253

Learn more about Calculus: brainly.com/question/27805589

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

3 0

1 year ago