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3 years ago

13

7 2/3 - 5 5/8how to solve

2 answers:

Sergeu [11.5K]3 years ago

8 0

First change to fractions   7 2/3 = 23/3      5 5/8 = 45/8

now you have 23/3 - 45/8     find least common denominator...24

23/3 = 184/24 -        45/8 =135/24 now your prob is 184/24 - 135/24

49/24   then change back 2 1/24

Lyrx [107]3 years ago

6 0

Change the numbers into improper fractions then subtract

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Determine two numbers that have a sum of 12 but a difference of 12

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Answer:

0 And 12

Step-by-step explanation:

0 + 12 = 12

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The function f(x) = x2 + 6x + 3 is transformed such that g(x) = f(x − 2). Find the vertex of g(x).

garri49 [273]

Answer:

(-1,-6)

Step-by-step explanation:

$f(x)=x^2+6x+3$

$g(x)=f(x-2)$

$g(x)=(x-2)^2+6(x-2)+3$ (Replaced x in f with (x-2))

$g(x)=x^2-4x+4+6x-12+3$ (Used $(x+b)^2=x^2+2bx+b^2$ and distributive property)

$g(x)=x^2-4x+6x+4-12+3$ (Gathered like terms)

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$\frac{-2}{2(1)}=-1$

Now we can obtain $g(\frac{-b}{2a})$ which is $g(-1)$ in this case:

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3 years ago

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2 years ago

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:

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

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

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

4 0

3 years ago