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2 years ago

What law can be used to determine that the conclusion is valid based on the given statements

Mathematics

1 answer:

ryzh [129]2 years ago

8 0

Answer:

A. Law of detachment

Step-by-step explanation:

The Law of detachment implies that when one condition is fulfilled the other cannot be and vice versa, then it is made the conclusion.

This condition is made the conclusion.

The Acute and Obtuse are detached of each other.

The acute angle is one in which the value of the angle is less than 90 degrees and obtuse angle is one in which the angle is greater than 90 degrees but less than 180 degrees.

Thus angles less than 90 degrees are acute and greater than 90 degrees are obtuse.

The conclusion of the given statement is valid based on the law of detachment as the condition has been made a conclusion.

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Match them together to correct answer

Verizon [17]

ANSWER

Frequency=2

Midline =3

Amplitude =5

Period =π

EXPLANATION

We have:

$f(x) = 5 \cos(x) + 3$

The given function is of the form;

$f(x) = a\cos(bx) + c$

where

a=|5| =5 is the amplitude

and b=2 is the frequency.

and 2π/b=2π/2=π

The range of the given function is

8≤y≤-2

The midline is

$\frac{8 + - 2}{2} = \frac{6}{2} = 3$

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3 years ago

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Which of the following are solutions to the equation below?

sladkih [1.3K]

Answer:

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

Step-by-step explanation:

considering the equation

$2x^2\:-\:4x\:-\:3\:=\:x$

solving

$2x^2\:-\:4x\:-\:3\:=\:x$

$2x^2-4x-3-x=x-x$

$2x^2-5x-3=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=-5,\:c=-3:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

solving

$x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5+\sqrt{49}}{2\cdot \:2}$

$x=\frac{5+7}{4}$

$x=3$

also solving

$x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5-\sqrt{49}}{4}$

$x=-\frac{2}{4}$

$x=-\frac{1}{2}$

Therefore,

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

7 0

3 years ago

Answer? I need it for math.

ladessa [460]

Answer:

The answer is D i guess . ( Not completely sure )

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2 years ago

You can rent time on computers at the local copy center for a a $ 7 setup charge and an additional $ 3.50 for every 10 minutes

bazaltina [42]

Answer:

The answer should round up to 40 minutes. Do it to make sure the answer is 4 which in this case = 40

Step-by-step explanation:

$22 - 7 \div 3.50$

8 0

2 years ago

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What error did Clyde make? And what is the correct simplified form

romanna [79]

Answer:

Clyde added -8+(-2), your supposed to multiply -8*(-2) the answer is j^16

Step-by-step explanation:

3 0

3 years ago

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