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3 years ago

To solve for m in the formula , use the multiplication property of equality. 1)True or 2) False?

Mathematics

2 answers:

VLD [36.1K]3 years ago

4 0

Answer:

the answer is true thanks to whoever gave the formula.

have a good day

Step-by-step explanation:

NeX [460]3 years ago

3 0

True. just like Texaschic101 said DV= M

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3 years ago

How do i solve this question

frozen [14]

$\bf \begin{cases}
h(x)=(f\circ g)(x)\\
h(x)=\sqrt{x+5}\\
f(x)=\sqrt{x+2}
\end{cases}\\\\
-------------------------------\\\\
(f\circ g)(x)\implies f[\ g(x)\ ]\implies f[\ g(x)\ ]=\sqrt{g(x)+2}\qquad thus
\\\\\\
\sqrt{g(x)+2}=(f\circ g)(x)=h(x)=\sqrt{x+5}\qquad therefore
\\\\\\
\sqrt{g(x)+2}=\sqrt{x+5}\impliedby \textit{squaring both sides}\\\\\\
g(x)+2=x+5\implies g(x)=x+5-2\implies \boxed{g(x)=x+3}$

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3 years ago

Please help solve for x..

aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

$\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0$

Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

Zero Product Property: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the Zero Product Property, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$

Therefore, the solutions to the given equation are: x = 1 or x = -1

Learn more here:

brainly.com/question/27751281

brainly.com/question/21186424

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1 year ago