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4 years ago

The sum of two numbers is 60. The larger number is 18 more than the smaller number. What are the numbers?

Mathematics

1 answer:

ASHA 777 [7]4 years ago

6 0

The two numbers are 21 and 39

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Which of the following is not a correct name for the line shown?

Talja [164]

Answer:

Step-by-step explanation:

The line can be named with small latin letters or with two large latin letters which represent two points lying on the line.

Consider all options:

A. Points V and X lie on the line, so you can name the line as VX (true option)

B. Points W and X lie on the line, so you can name the line as WX (true option)

C. p represents the line (small letter given on the diagram) - (true option)

D. You cannot name the line using point V and line p, so this option is false

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3 years ago

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Which postulate explains why a tripod will not tip over?

Feliz [49]

Tripod will not tip over because: a plane contains at least 3 noncollinear points .
Answer: B )

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3 years ago

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Last one, use trigonometry to help you solve for y given that one side is 4 and there is an angle of 50 degrees.

Ostrovityanka [42]

Answer:

y = 6.22

You can solve this in two ways.

1.) Use SOH CAH TOA:

I typically start off by labeling the sides of the triangle with H (hypotenuse), O (opposite), and A (adjacent). Because I need to figure out what y is when given an angle and 4, I will use CAH, or cosine.

$\cos(angle) = \frac{adjacent}{hypotenuse}$

$\cos(50) = \frac{4}{y}$

$\cos(50) \times y = 4$

$y = \frac{4}{ \cos(50) }$

$y = 6.22$

2.) Use Law of Sines:

Solve for the last angle inside the triangle first.

$180 - (50 + 90) = 40$

Then use the angle you found (40°) in the equation.

$\frac{4}{ \sin(40) } = \frac{y}{ \sin(90) }$

$\frac{4}{ \sin(40) } \times \sin(90) = y$

$y = 6.22$

3 0

3 years ago

Find the Nth term of this number sequnce<br> 7,10,13,16

Strike441 [17]

$\text{It is an arithmetic series.}\\\\\text{First term,}~ a = 7\\\\\text{Common difference,}~ d=10 -7 = 3\\\\\text{Nth term} = a + (n -1) d\\\\~~~~~~~~~~~~~=7+(n-1)3\\\\~~~~~~~~~~~~~=7+3n-3\\\\~~~~~~~~~~~~~=4+3n$

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3 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

2 years ago