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2 years ago

11

I though of a number tripple it and add fifteen the result is 45. What is the number i thought of?

2 answers:

Sholpan [36]2 years ago

6 0

Answer:

10

Step-by-step explanation:

STALIN [3.7K]2 years ago

5 0

Answer:

10

Step-by-step explanation:

because 10 times three is 30 plus 15 equals 45

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If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalen

larisa [96]

What are your answer choices? I’m not sure which equation they are asking for?

The solution to this system of equations is
X= -5
Y= -3

4 0

3 years ago

Drake produced 52 songs last year. If each new album has 15 songs on it, how many of his songs were not released on an album?

Andreyy89

Answer:

7

Step-by-step explanation:

If each album has 15 songs, 15 * 3 = 45 so 52 - 45 = 7. Therefore, Drake produced 7 songs not in an album.

7 0

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

Kamala spent a total of $4.32 to buy

Ostrovityanka [42]

Each pencil costs 0.72 as you do 4.32 / 6

7 0

2 years ago

For the following pair of functions, find (f+g)(x) and (f-g)(x).

ch4aika [34]

Given:

The functions are

$f(x)=4x^2+7x-5$

$g(x)=-9x^2+4x-13$

To find:

The functions $(f+g)(x)$ and $(f-g)(x)$ .

Solution:

We know that,

$(f+g)(x)=f(x)+g(x)$

$(f+g)(x)=4x^2+7x-5-9x^2+4x-13$

$(f+g)(x)=(4x^2-9x^2)+(7x+4x)+(-5-13)$

$(f+g)(x)=-5x^2+11x-18$

And,

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

$(f-g)(x)=(4x^2+7x-5)-(-9x^2+4x-13)$

$(f+g)(x)=4x^2+7x-5+9x^2-4x+13$

$(f+g)(x)=(4x^2+9x^2)+(7x-4x)+(-5+13)$

$(f-g)(x)=13x^2+3x+8$

Therefore, the required functions are $(f+g)(x)=-5x^2+11x-18$

and $(f-g)(x)=13x^2+3x+8$ .

7 0

3 years ago