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3 years ago

For every 6 boxes sold, Stephanie makes a profit of $8.10. Which table shows this same rate of change?

Mathematics

1 answer:

kumpel [21]3 years ago

4 0

The answer is A
The answer is A

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Zack and Noah jogged 2/5 of a mile and walked another 1/4 of a mile. How far did they go in all?

Basile [38]

Find a common denominator. The lowest one would come from you multiplying 2/5 by 4 on each side and 1/4 by 5 on each side. You are left with 8/20 and 5/20, added together would be 13/20.

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2 years ago

How do you use the GCF to write equivalent fractions?

Anastasy [175]

Answer:

Step-by-step explanation:

To find equivalent fractions, we multiply the numerator and the denominator by the same number, so we need to multiply the denominator of 7 by a number that will give us 21. Since 3 multiplied by 7 gives us 21, we can find an equivalent fraction by multiplying both the numerator and denominator by 3.Oct 4, 2021

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2 years ago

Question in pictures

yan [13]

The derivatives of the functions are listed below:

(a) $f'(x) = -7\cdot x^{-\frac{9}{2} }- 2\cdot x + 4 - \frac{1}{5} - 5\cdot x^{-2}$

(b) $f'(x) = \frac{1}{3}\cdot (x + 3)^{-\frac{2}{3} }\cdot (x+ 5)^{\frac{1}{3} } + \frac{1}{3} \cdot (x + 5)^{-\frac{2}{3} } \cdot (x + 3)^{\frac{1}{3} }$

(c) f'(x) = [(cos x + sin x) · (x² - 1) - (sin x - cos x) · (2 · x)] / (x² - 1)²

(d) f'(x) = (5ˣ · ㏑ 5) · ㏒₅ x + 5ˣ · [1 / (x · ㏑ 5)]

(e) f'(x) = 45 · (x⁻⁵ + √3)⁻⁸ · x⁻⁶

(f) $f'(x) = (\ln x + 1)\cdot [7^{x\cdot \ln x \cdot \ln 7}+7\cdot (x\cdot \ln x)^{6}]$

(g) $f'(x) = -2\cdot \arccos x \cdot \left(\frac{1}{\sqrt{1 - x^{2}}} \right) - \left(\frac{1}{1 + x} \right) \cdot \left(\frac{1}{2} \cdot x^{-\frac{1}{2} }\right)$

(h) f'(x) = cot x + cos (㏑ x) · (1 / x)

<h3>How to find the first derivative of a group of functions</h3>

In this question we must obtain the <em>first</em> derivatives of each expression by applying <em>differentiation</em> rules:

(a) $f(x) = 2 \cdot x^{-\frac{7}{2} } - x^{2} + 4 \cdot x - \frac{x}{5} + \frac{5}{x} - \sqrt[11]{2022}$

$f(x) = 2 \cdot x^{-\frac{7}{2} } - x^{2} + 4 \cdot x - \frac{x}{5} + \frac{5}{x} - \sqrt[11]{2022}$ Given
$f(x) = 2 \cdot x^{-\frac{7}{2} } - x^{2} + 4\cdot x - \frac{x}{5} + 5 \cdot x^{-1} - \sqrt[11]{2022}$ Definition of power
$f'(x) = -7\cdot x^{-\frac{9}{2} }- 2\cdot x + 4 - \frac{1}{5} - 5\cdot x^{-2}$ Derivative of constant and power functions / Derivative of an addition of functions / Result

(b) $f(x) = \sqrt[3]{x + 3} \cdot \sqrt[3]{x + 5}$

$f(x) = \sqrt[3]{x + 3} \cdot \sqrt[3]{x + 5}$ Given
$f(x) = (x + 3)^{\frac{1}{3} }\cdot (x + 5)^{\frac{1}{3} }$ Definition of power
$f'(x) = \frac{1}{3}\cdot (x + 3)^{-\frac{2}{3} }\cdot (x+ 5)^{\frac{1}{3} } + \frac{1}{3} \cdot (x + 5)^{-\frac{2}{3} } \cdot (x + 3)^{\frac{1}{3} }$ Derivative of a product of functions / Derivative of power function / Rule of chain / Result

(c) f(x) = (sin x - cos x) / (x² - 1)

f(x) = (sin x - cos x) / (x² - 1) Given
f'(x) = [(cos x + sin x) · (x² - 1) - (sin x - cos x) · (2 · x)] / (x² - 1)² Derivative of cosine / Derivative of sine / Derivative of power function / Derivative of a constant / Derivative of a division of functions / Result

(d) f(x) = 5ˣ · ㏒₅ x

f(x) = 5ˣ · ㏒₅ x Given
f'(x) = (5ˣ · ㏑ 5) · ㏒₅ x + 5ˣ · [1 / (x · ㏑ 5)] Derivative of an exponential function / Derivative of a logarithmic function / Derivative of a product of functions / Result

(e) f(x) = (x⁻⁵ + √3)⁻⁹

f(x) = (x⁻⁵ + √3)⁻⁹ Given
f'(x) = - 9 · (x⁻⁵ + √3)⁻⁸ · (- 5) · x⁻⁶ Rule of chain / Derivative of sum of functions / Derivative of power function / Derivative of constant function
f'(x) = 45 · (x⁻⁵ + √3)⁻⁸ · x⁻⁶ Associative and commutative properties / Definition of multiplication / Result

(f) $f(x) = 7^{x\cdot \ln x} + (x \cdot \ln x)^{7}$

$f(x) = 7^{x\cdot \ln x} + (x \cdot \ln x)^{7}$ Given
$f'(x) = 7^{x\cdot\ln x} \cdot \ln 7 \cdot (\ln x + 1) + 7\cdot (x\cdot \ln x)^{6}\cdot (\ln x + 1)$ Rule of chain / Derivative of sum of functions / Derivative of multiplication of functions / Derivative of logarithmic functions / Derivative of potential functions
$f'(x) = (\ln x + 1)\cdot [7^{x\cdot \ln x \cdot \ln 7}+7\cdot (x\cdot \ln x)^{6}]$ Distributive property / Result

(g) $f(x) = \arccos^{2} x - \arctan (\sqrt{x})$

$f(x) = \arccos^{2} x - \arctan (\sqrt{x})$ Given
$f'(x) = -2\cdot \arccos x \cdot \left(\frac{1}{\sqrt{1 - x^{2}}} \right) - \left(\frac{1}{1 + x} \right) \cdot \left(\frac{1}{2} \cdot x^{-\frac{1}{2} }\right)$ Derivative of the subtraction of functions / Derivative of arccosine / Derivative of arctangent / Rule of chain / Derivative of power functions / Result

(h) f(x) = ㏑ (sin x) + sin (㏑ x)

f(x) = ㏑ (sin x) + sin (㏑ x) Given
f'(x) = (1 / sin x) · cos x + cos (㏑ x) · (1 / x) Rule of chain / Derivative of sine / Derivative of natural logarithm /Derivative of addition of functions
f'(x) = cot x + cos (㏑ x) · (1 / x) cot x = cos x / sin x / Result

To learn more on derivatives: brainly.com/question/23847661

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7 0

1 year ago

Does anyone know the answer

matrenka [14]

Answer:

Step-by-step explanation:

Try to understand what this equation is saying and what plugging in different values would represent.

At x=0, b(0)=850; which means the initial balance is $850. So H is incorrect.

At x=1, b(1)=871.25; which means that after 1 year, the initial balance will have increase to $871.25. So J is incorrect.

Since the initial vacation of 850 is being multiplied by a factor which is greater than one, the balance will be increased each year. So F is incorrect.

Finally, if we look at the factor by which we are multiplying, do a simple Algebraic step, and convert it into percentages we get:

1.025 = (1 + 0.025) = 100% + 2.5%

Essentially this is showing us that the balance will increase by 2.5% on however much is in the account each year.

So G is your answer.

8 0

3 years ago

The length of a rectangle is 9 less than 4 less than the width. The perimeter is 143. Find the width and length.

damaskus [11]

The formula for perimeter is 2l+2w. Since we know l=(w-4)-9, we can substitute this into the formula.

l=w-13

2(w-13)+2w=143

2w-26+2w=143

4w=169

w=42.25 units

Hope this helps!

6 0

2 years ago