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2 years ago

Given the function g(x)=√x+4, which of the following values cannot be the domain of g?

Mathematics

1 answer:

Gennadij [26K]2 years ago

6 0

Answer:

D. x = -5

Step-by-step explanation:

The square root function is only defined for non-negative arguments. For the function ...

$g(x)=\sqrt{x+4}$

this means we must have ...

x +4 ≥ 0

x ≥ -4 . . . . . . subtract 4 from both sides

All of the numbers on your answer choice list are more than -4 except -5.

x = -5 is not in the domain of g(x)

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The product of a binomial and trinomial is x^3+3x^2-x+2x^2+6x-2 which expression is equivalent to this product after it has been

Debora [2.8K]

Answer: First option.

Step-by-step explanation:

You know that the product obtained by multiplying a binomial and a trinomial is:

$x^3+3x^2-x+2x^2+6x-2$

Then, in order to simplify this product, it is necessary to add the like terms.

Therefore, the expression that is equivalent to this product after it has been fully simplified is:

$=x^3+(3x^2+2x^2)+(6x-x)-2\\\\=x^3+5x^2+5x-2$

You can observe that this matches with the first option.

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2 years ago

Please help with this question asap i will give brainliest

sergiy2304 [10]

Answer:

B is your answer,

With the factors of ( 2 )( 1 ), the lay out of the equation makes only sense with answer B.

6 0

3 years ago

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From the chapter Rational Numberssome body please answer fast

Alexxx [7]

Answer:

- $\frac{3}{7}$

Step-by-step explanation:

Given

- $\frac{42}{98}$ ← divide numerator/ denominator by 14 )

= - $\frac{3}{7}$

7 0

2 years ago

Solve the equation 2cosA = 3tanA

zavuch27 [327]

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\
-----------------------------\\\\

2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)}
\\\\\\
2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A)
\\\\\\
2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2$

$\bf \\\\\\
0=[2sin(A)-1][sin(A)+2]\implies 
\begin{cases}
0=2sin(A)-1\\
1=2sin(A)\\
\frac{1}{2}=sin(A)\\\\
sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\
\frac{\pi }{6},\frac{5\pi }{6}\\
----------\\
0=sin(A)+2\\
-2=sin(A)
\end{cases}$

now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1

now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine

so, only the first case are the valid angles for A

8 0

2 years ago

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a_sh-v [17]

Answer:

20 bottles

Step-by-step explanation:

Number of bottles

$= 5 \div \frac{1}{4} \\ \\ = 5 \times \frac{4}{1} \\ \\ = 20$

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2 years ago

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