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3 years ago

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What is the value of x?

1 answer:

Over [174]3 years ago

4 0

Answer:

c-12

Step-by-step explanation:

Pythagoras theorem states that a²+b²=c²

144+b²=c²

b²=c²-144

b=c-12

b=x

x=c-12

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Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

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erastovalidia [21]

Answer:

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Step-by-step explanation:

The area of the triangle = $16 cm^2$

Base of the triangle = 10 cm

Height of the triangle = h cm

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$16 cm^2=\frac{1}{2}\times 10 cm \times h$

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8 0

3 years ago

Read 2 more answers

dalvyx [7]

So,

All we have to do is multiply all three dimensions together.

First, convert all fractions into improper fractions.

$3 \frac{1}{5}--\ \textgreater \ \frac{16}{5}$

$1 \frac{3}{4} --\ \textgreater \ \frac{7}{4}$

$2 \frac{1}{4} --\ \textgreater \ \frac{9}{4}$

Now multiply the fractions together.

<span> $\frac{16}{5}*\frac{7}{4}*\frac{9}{4}= \frac{2*2*2*2*3*3*7}{2*2*2*2*5}$
</span>
$\frac{3*3*7}{5}--\ \textgreater \ \frac{63}{5}\ or\ 12 \frac{3}{5}\ ft.^3$

The correct option is C.

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3 years ago

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Lera25 [3.4K]

Answer:5

Step-by-step explanation:

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