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3 years ago

Need help. How do you start?

Mathematics

1 answer:

Aloiza [94]3 years ago

7 0

Answer:

it is a right angle triangle therefore

h²=b²+p²

8²=4²+p²

64-16=p²

48=p²

under root 48 =p

6.928

so 6.93 is the answer

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4 years ago

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lidiya [134]

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3 years ago

The diameter of a circle is 18 cm. Find its area in terms of pi

Kobotan [32]

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Step-by-step explanation:

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3 years ago

The smallest object visible with your eyes is similar to the width of a piece of hair, which is 1×10−4 meters wide. Using an opt

Sloan [31]

Answer:

B. $5\times10^{2}$

Step-by-step explanation:

We are told that the smallest object visible with our eyes is similar to the width of a piece of hair, which is $1\times 10^{-4}$ meters wide.

Using an optical microscope, we can see items up to $2\times 10^{-7}$ meters wide.

To find the objects we can see with our eyes are how much larger than the objects we can see with an optical microscope, we can set an equation as:

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=\frac{1*10^{-4}}{2*10^{-7}}$

Using the exponent rule of quotient $\frac{a^m}{a^n}=a^{m-n}$ we will get,

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=\frac{1}{2}*10^{-4-(-7)}$

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10^{-4+7}$

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10^{3}$

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10\times 10^{3-1}$

$\text{The object we can see with our eyes}=5\times10^{2}*\text{The objects we can see with microscope}$

Therefore, the objects we can see with our eyes are $5\times10^{2}$ times larger than the objects we can see with an optical microscope and option B is the correct choice.

3 0

3 years ago

Help me pls! Thank you so much

frosja888 [35]

$\bf cos\left[tan^{-1}\left(\frac{12}{5} \right)+ tan^{-1}\left(\frac{-8}{15} \right) \right]\\
\left. \qquad \qquad \quad \right.\uparrow \qquad \qquad \qquad \uparrow \\
\left. \qquad \qquad \quad \right.\alpha \qquad \qquad \qquad \beta
\\\\\\
\textit{that simply means }tan(\alpha)=\cfrac{12}{5}\qquad and\qquad tan(\beta)=\cfrac{-8}{5}
\\\\\\
\textit{so, we're really looking for }cos(\alpha+\beta)$

now.. hmmm -8/15 is rather ambiguous, since the negative sign is in front of the rational, and either 8 or 15 can be negative, now, we happen to choose the 8 to get the minus, but it could have been 8/-15

ok, well hmm so, the issue boils down to

$\bf tan(\theta)=\cfrac{opposite}{adjacent}\qquad thus
\\\\\\
tan(\alpha)=\cfrac{12}{5}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a}
\\\\\\
\textit{so, what is the hypotenuse "c"?}\\
\textit{ well, let's use the pythagorean theorem}
\\\\\\
c=\sqrt{a^2+b^2}\implies c=\sqrt{25+144}\implies c=\sqrt{169}\implies \boxed{c=13}\\\\
-----------------------------\\\\
\textit{this simply means }\boxed{cos(\alpha)=\cfrac{5}{13}\qquad \qquad sin(\alpha)=\cfrac{12}{13}
}$

now, let's take a peek at the second angle, angle β

$\bf tan(\beta)=\cfrac{-8}{15}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a}
\\\\\\
\textit{again, let's find "c", or the hypotenuse}
\\\\\\
c=\sqrt{15^2+(-8)^2}\implies c=\sqrt{289}\implies \boxed{c=17}\\\\
-----------------------------\\\\
thus\qquad \boxed{cos(\beta)=\cfrac{15}{17}\qquad \qquad sin(\beta)=\cfrac{-8}{17}}$

now, with that in mind, let's use the angle sum identity for cosine

$\bf cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}})\\\\
-----------------------------\\\\
cos({{ \alpha}} + {{ \beta}})= \left( \cfrac{5}{13} \right)\left( \cfrac{15}{17} \right)-\left( \cfrac{12}{13} \right)\left( \cfrac{-8}{17} \right)
\\\\\\
cos({{ \alpha}} + {{ \beta}})= \cfrac{75}{221}-\cfrac{-96}{221}\implies cos({{ \alpha}} + {{ \beta}})= \cfrac{75}{221}+\cfrac{96}{221}
\\\\\\
\boxed{cos({{ \alpha}} + {{ \beta}})=\cfrac{171}{221}}$

8 0

3 years ago