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2 years ago

9

Abigail is driving on a long road trip. She currently has 13 gallons of gas in her car. Each hour that she drives, her car uses

up 1 gallon of gas. How much gas would be in the tank after driving for 2 hours? How much gas would be left after tt hours?

1 answer:

denpristay [2]2 years ago

6 0

Answer:

11

Step-by-step explanation:

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You are buying tickets to a school event. You may purchase at most 8 tickets to receive a discount. Write an inequality to expre

podryga [215]

8 is greater than or equal to x
(where x = the number of tickets)

or x is less than or equal to 8

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3 years ago

Merina is scheduled to make two loan payments to Bradford in the amount of $1,000 each, two months and nine months from now. Mer

Svetach [21]

well, we're assuming all along that Merina owes Bradford $2000, because in the 1st scenario, she was going to pay twice $1000.

on the 2nd scenario, she'll be paying the same $2000 but split 7 months from now and then 7 months later, same 2000 bucks, at which point Bradford applied 8.5% interest.

using those assumptions, since the wording is not quite clear, we can say that Merina is simply paying 2000 bucks plus the 8.5%

$\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{8.5\% of 2000}}{\left( \cfrac{8.5}{100} \right)2000}\implies 170 \\\\[-0.35em] ~\dotfill\\\\ \cfrac{\stackrel{principal}{2000}~~ + ~~\stackrel{interest}{170}}{2}\implies \stackrel{\textit{two equal payments of}}{1085}$

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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:

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

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

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

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3 years ago

A week before the election, 1,500 people were asked who they planned to vote for. 45% said they planned to vote for Candidate X.

Nastasia [14]

The answer to you question is 17% Of the people have not yet decided

3 0

3 years ago

Complete the factorization of 3x ^ 2 - 10x + 8 3x^ 2 -10x+8=(x- Box x-4)

aniked [119]

Answer:

$3x^2-10x+8=(x-x_1)(x-x_2)=(x-2)(x-\frac{4}{3})$

Step-by-step explanation:

The general form of a quadratic polynomial is given by:

$ax^2+bx+c$ (1)

You have the following polynomial:

$3x^2-10x+8$ (2)

In order to complete the factorization you can use the quadratic formula, to obtain the roost of the polynomial. The quadratic formula is given by:

$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (3)

By comparing the equation (1) with the equation (2) you obtain:

a = 3

b = -10

c = 8

Then, you replace these values in the equation (3):

$x_{1,2}=\frac{-(-10)\pm \sqrt{(-10)^2-4(3)(8)}}{2(3)}\\\\x_{1,2}=\frac{10\pm2}{6}\\\\x_1=2\\\\x_2=\frac{4}{3}$

Then, the factorization of the polynomial is:

$3x^2-10x+8=(x-x_1)(x-x_2)=(x-2)(x-\frac{4}{3})$

5 0

2 years ago