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2 years ago

Can someone help me please?

Mathematics

1 answer:

Mars2501 [29]2 years ago

8 0

Answer:

Answer below

Step-by-step explanation:

Point D: (-6,-1)

Point C: (-3,0)

Point B: (-3,6)

Point A: (0,0)

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Algebra: Find x<br> ....

luda_lava [24]

Answer:

Step-by-step explanation:

The triangles are similar, thus the ratios of corresponding sides are equal, that is

$\frac{2x+1}{x+4}$ = $\frac{40}{25}$ = $\frac{8}{5}$ ( cross- multiply )

5(2x + 1) = 8(x + 4) ← distribute parenthesis on both sides )

10x + 5 = 8x + 32 ( subtract 8x from both sides )

2x + 5 = 32 ( subtract 5 from both sides )

2x = 27 ( divide both sides by 2 )

x = 13.5 → d

3 0

4 years ago

What pattern can be observed from the data?

marin [14]

I can’t see the diagram

5 0

3 years ago

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What is the domain of the function y=tan (x/8)<br> Please Help<br> 15 Points

7nadin3 [17]

1. You have the function $y=tan(\frac{x}{8} )$

2. You know that $y=\frac{sinx}{cosx}$ with $cosx\neq 0$ , because the division by zero is not defined. So , $cosx=0$ for $\frac{(2n+1)\pi}{2}$ where $n$ is a integer number.

3. Then, to find the domain of the function given in the problem, you must make $\frac{x}{8} \neq \frac{k\pi}{2} \\ x\neq 4k\pi$

Where $k$ is odd number.

4. Therefore:

$x\neq 4\pi (2n+1)$

5. Finally, the domain is: $4\pi (2n+1)$

The answer is: $4\pi (2n+1)$

5 0

3 years ago

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Factor completely <br> 4(x+1)^2/3 + 12(x+1)^-1/3

wolverine [178]

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
\left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}
\\\\
-------------------------------\\\\
%4(x+1)^2/3 + 12(x+1)^-1/3
4(x+1)^{\frac{2}{3}}+12(x+1)^{-\frac{1}{3}}\implies 4(x+1)^{\frac{2}{3}}+12\cfrac{1}{(x+1)^{\frac{1}{3}}}
\\\\\\$

$\bf 4(x+1)^{\frac{2}{3}}+\cfrac{12}{(x+1)^{\frac{1}{3}}}\impliedby \textit{so, our LCD is }(x+1)^{\frac{1}{3}}
\\\\\\
\cfrac{4(x+1)^{\frac{2}{3}}\cdot (x+1)^{\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)^{\frac{2}{3}+\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}
\\\\\\
\cfrac{4(x+1)^{\frac{3}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4x+4+12}{(x+1)^{\frac{1}{3}}}
\\\\\\
\cfrac{4x+16}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+4)}{\sqrt[3]{x+1}}$

3 0

3 years ago

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Is the equation 4y+6=5y-8 true,false,or open

svlad2 [7]

True.

Work is in attached sheet! <3

5 0

3 years ago