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3 years ago

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Abigail's and gina's ages are consecutive integers. abigail is younger than gina and gina's age is represented by x. if the diff

erence of the square of gina's age and eight times abigail's age is 17, which equation could be used to fi nd gina's age?

1 answer:

Ostrovityanka [42]3 years ago

4 0

If Gina's age is x and Abigail is younger than Gina, Abigail's age will be represented by x-1 given that their ages are consecutive integers.

Translating the fact that the the difference of the square of Gina's age and eight times Abigail's age is 17, we will have the following equation:

$x^{2} -8(x-1)=17$

We can use the equation above to find Gina's age.

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Please answer this question, i request

Jet001 [13]

${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

$\star \: \tt \cot \theta = \dfrac{7}{8}$

${\large{\textsf{\textbf{\underline{\underline{To \: Evaluate :}}}}}}$

$\star \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) }$

${\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}$

Consider a $\triangle$ ABC right angled at C and $\sf \angle \: B = \theta$

Then,

‣ Base [B] = BC

‣ Perpendicular [P] = AC

‣ Hypotenuse [H] = AB

$\therefore \tt \cot \theta = \dfrac{Base}{ Perpendicular} = \dfrac{BC}{AC} = \dfrac{7}{8}$

Let,

Base = 7k and Perpendicular = 8k, where k is any positive integer

In $\triangle$ ABC, H² = B² + P² by Pythagoras theorem

$\longrightarrow \tt {AB}^{2} = {BC}^{2} + {AC}^{2}$

$\longrightarrow \tt {AB}^{2} = {(7k)}^{2} + {(8k)}^{2}$

$\longrightarrow \tt {AB}^{2} = 49{k}^{2} + 64{k}^{2}$

$\longrightarrow \tt {AB}^{2} = 113{k}^{2}$

$\longrightarrow \tt AB = \sqrt{113 {k}^{2} }$

$\longrightarrow \tt AB = \red{ \sqrt{113} \: k}$

Calculating Sin $\sf \theta$

$\longrightarrow \tt \sin \theta = \dfrac{Perpendicular}{Hypotenuse}$

$\longrightarrow \tt \sin \theta = \dfrac{AC}{AB}$

$\longrightarrow \tt \sin \theta = \dfrac{8 \cancel{k}}{ \sqrt{113} \: \cancel{ k } }$

$\longrightarrow \tt \sin \theta = \purple{ \dfrac{8}{ \sqrt{113} } }$

Calculating Cos $\sf \theta$

$\longrightarrow \tt \cos \theta = \dfrac{Base}{Hypotenuse}$

$\longrightarrow \tt \cos \theta = \dfrac{BC}{ AB}$

$\longrightarrow \tt \cos \theta = \dfrac{7 \cancel{k}}{ \sqrt{113} \: \cancel{k } }$

$\longrightarrow \tt \cos \theta = \purple{ \dfrac{7}{ \sqrt{113} } }$

<u>Solving the given expression</u><u> </u><u>:</u><u>-</u><u> </u>

$\longrightarrow \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) }$

Putting,

• Sin $\sf \theta$ = $\dfrac{8}{ \sqrt{113} }$

• Cos $\sf \theta$ = $\dfrac{7}{ \sqrt{113} }$

$\longrightarrow \: \tt \dfrac{ \bigg(1 + \dfrac{8}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{8}{ \sqrt{133}} \bigg) }{\bigg(1 + \dfrac{7}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{7}{ \sqrt{133}} \bigg)}$

<u>Using</u><u> </u><u>(</u><u>a</u><u> </u><u>+</u><u> </u><u>b</u><u> </u><u>)</u><u> </u><u>(</u><u>a</u><u> </u><u>-</u><u> </u><u>b</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>a²</u><u> </u><u>-</u><u> </u><u>b²</u>

$\longrightarrow \: \tt \dfrac{ { \bigg(1 \bigg)}^{2} - { \bigg( \dfrac{8}{ \sqrt{133} } \bigg)}^{2} }{ { \bigg(1 \bigg)}^{2} - { \bigg( \dfrac{7}{ \sqrt{133} } \bigg)}^{2} }$

$\longrightarrow \: \tt \dfrac{1 - \dfrac{64}{113} }{ 1 - \dfrac{49}{113} }$

$\longrightarrow \: \tt \dfrac{ \dfrac{113 - 64}{113} }{ \dfrac{113 - 49}{113} }$

$\longrightarrow \: \tt { \dfrac { \dfrac{49}{113} }{ \dfrac{64}{113} } }$

$\longrightarrow \: \tt { \dfrac{49}{113} }÷{ \dfrac{64}{113} }$

$\longrightarrow \: \tt \dfrac{49}{ \cancel{113}} \times \dfrac{ \cancel{113}}{64}$

$\longrightarrow \: \tt \dfrac{49}{64}$

$\qquad \: \therefore \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) } = \pink{\dfrac{49}{64} }$

$\begin{gathered} {\underline{\rule{300pt}{4pt}}} \end{gathered}$

${\large{\textsf{\textbf{\underline{\underline{We \: know :}}}}}}$

✧ Basic Formulas of Trigonometry is given by :-

$\begin{gathered}\begin{gathered}\boxed { \begin{array}{c c} \\ \bigstar \: \sf{ In \:a \:Right \:Angled \: Triangle :} \\ \\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star \tan \theta = \dfrac{Perpendicular}{Base}}\\\\ \sf{\star \cosec \theta = \dfrac{Hypotenuse}{Perpendicular}} \\\\ \sf{\star \sec \theta = \dfrac{Hypotenuse}{Base}}\\\\ \sf{\star \cot \theta = \dfrac{Base}{Perpendicular}} \end{array}}\\\end{gathered} \end{gathered}$

${\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}$

✧ Figure in attachment

$\begin{gathered} {\underline{\rule{200pt}{1pt}}} \end{gathered}$

3 0

2 years ago

Image attachment below, Algebra 1 stuff

chubhunter [2.5K]

I know you didn't want an explanation, but I'll give you one anyway.
So, for a function, when you have f(5) = , then you'll put that 5 in for all of the x's in the function.

The first batch of problems gave you the equation f(x) = $\frac{x}{2}$ + 3. Now, plug in the numbers given for s and solve.

1. f(5) = <span> $\frac{x}{2}$ + 3 Plug in 5 for x
</span> = $\frac{5}{2}$ + 3 Make three into a fraction over 2
= $\frac{5}{2}$ + <span> $\frac{6}{2}$ Add
= </span><span><span> $\frac{11}{2}$

</span>2. f(-4) = </span><span> $\frac{x}{2}$ + 3 Plug in -4 for x
= </span><span> $\frac{-4}{2}$ + 3 Divide -4 by 2
= -2 + 3 Add
= 1

For the next set, you have g(x) = x</span>² + 1.

3. Let's split this up. Solve the first equation, then the second, and then put the answers together to solve it.

g(4) = x² + 1 Plug in 4 for x
= 4² + 1 Square
= 16 + 1 Add
= 17

g(3) = x² + 1 Plug in 3 for x
= 3² + 1 Square
= 9 + 1 Add
= 10

Now, put the two together.

g(4) + g(3) = Substitute in the answers you just got.
17 + 10 = Add
= 27

4. g(-1) = x² + 1 Substitute in -1 for x
= (-1)² + 1 Square
= 1 + 1 Add
= 2

5. g(-6) = x² + 1 Plug in -6 for x
= (-6)² + 1 Square
= 36 + 1 Add
= 37

For the last set, we have h(x) = x² + 7 and k(x) = 4x - 5. We'll have to pay close attention to the starting variables here.

6. h(-6) = x² + 7 Plug in -6 for x
= (-6)² + 7 Square
= 36 + 7 Add
= 43

k(4) = 4x - 5 Plug in 4 for x
= 4(4) - 5 Multiply
= 16 - 5 Subtract
= 11

Put the two answers into the final equation.

h(-6) + k(4) = Substitute in the answers you got
43 + 11 = Add
= 54

7. k(10) = 4x - 5 Plug in 10 for x
= 4(10) - 5 Multiply
= 40 - 5 Subtract
= 35

h(2) = x² + 7 Plug in 2 for x
= 2² + 7 Square
= 4 + 7 Add
= 11

Now, put those into the equation.

k(10) - h(2) = Plug in the answers you got
35 - 11 = Subtract
= 24

8. For this one, it wants you to add h(x) and k(x). We don't need to plug anything into the equations this time because they're both already solved for x! So, you can just set this up like a standard find-x equation.

(x² + 7) + (4x - 5) =
x² + 7 + 4x - 5 = Combine like terms (7 and -5)
x² + 2 + 4x = Reorder so the variables come first
= x² + 4x + 2

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3 years ago

Read 2 more answers

The n sequence question

MaRussiya [10]

Answer:

a) in order

-3;-2;-1

b) 6

1-4=-3

2-4=-2

3-4=-1

4-4=0

5-4=1

6-4=2

7-4=3

8-4=4

9-4=5

10-4=6

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3 years ago

sea level =0 A submarine is 120ft below sea level it climbs 50 ft towards sea level where the submarine comparrf to sea level an

Katyanochek1 [597]

Answer:

<h2>70 ft</h2>

Step-by-step explanation:

120 below we can use a model to figure this out.

sea level is 0

the submarine <u>goes 120 down</u>, so its -120

it <u>goes up by 50</u>, so now its <em>70</em>

no other obstacles, 70 or A is your answer.

7 0

3 years ago

I'm solving systems and I don't understand how to do it pls helpp

kaheart [24]

Answer:

Use the process of elimination. I'll only do one example since you have 4.

Step-by-step explanation:

Let's do the first question.

2x-3y=-11

2x+y=9

Our goal here is to be left with one variable to solve, this way we can use the answer for that variable to find the other. A.K.A only find X to plug into the equation to find Y. To do this, we'll use the process of elimination, in which we need to cancel out a variable.

X seems the easiest here. To cancel 2 (in the top equation), you need to have a -2. And that -2 needs to be the x coefficient in the 2nd equation, in order for us to "cancel" it out. To get -2, you multiply 2 by -1.

-1(2x+y)= -1(9)

Note that whatever you do to one side, you need to do it to both sides! Hence, why I also multiplied 9 by -1. You should now be left with:

-2x -y = -9

Now, we cancel out the positive 2 and -2 from both equations by adding the two equations.

2x-3y= 11

+

-2x-y=-9

X cancels out, so we're left with -4y=-20. Solve for Y, which is 5.

-4y=-20

y=5

Now that you have the Y value, plug it into any of the two equations to find your X value.

2x-3y=11

2x-3(5)=11

2x-15=11

2x=26

x=13

And voilah! Our Y value is 5, and our X value is 13.

It's pretty simple once you understand what's going on. All that we did was to cancel out one variable (either X or Y) in the two equations so that we're left with only one variable to solve. If we cancel out X, we only need to solve for Y. You do this by manipulating one of the two equations (doesn't matter which one). Once you find either X or Y, all you need to is to plug it into one of the original equations. Then, you'll have both X and Y.

And that my friend, is the process of elimination. Good luck!

8 0

3 years ago

Read 2 more answers