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stiv31 [10]
2 years ago
10

Ella is putting her savings into a retirement fund that will compound her $7000 semiannually at 7% interest. How much will her r

etirement fund be worth after 10 years? *
Mathematics
1 answer:
forsale [732]2 years ago
6 0

Answer:

8099

Step-by-step explanation:

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What is "two more than the quotient of six and a number, n" written as an algebraic expression
Yuri [45]

Answer:

The algebraic expression of "two more than the quotient of six and a number, n is 2 + 6/n

Step-by-step explanation:

The quotient number of six and a number n = 6 / n

two more than the quotient of six and a number, n = 2 + 6/n

The algebraic expression of "two more than the quotient of six and a number, n is 2 + 6/n

4 0
3 years ago
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M
zlopas [31]
Not understanding what’s the question.
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2 years ago
Solve the equation -3(x - 14) + 9x = 6x + 42. Does the equation have one solution, no solution, or infinitely many solutions?
natulia [17]

Hello!

\large\boxed{\text{Infinitely many solutions.}}

-3(x - 14) + 9x = 6x + 42

Distribute the coefficient of the parenthesis:

-3(x) - 3(-14) + 9x = 6x + 42

-3x + 42 + 9x = 6x + 42

Combine like terms:

6x + 42 = 6x + 42

Both sides of the equation are the same, therefore:

There are infinitely many solutions.

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2 years ago
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Question 3 of 10
mestny [16]

Answer:

㋡

Check Answer

♣ Qᴜᴇꜱᴛɪᴏɴ :

If tan θ = \sf{\dfrac{1}{\sqrt{7}}}

7

1

, Show that \sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=

4

3

★═════════════════★

♣ ᴀɴꜱᴡᴇʀ :

We know :

\large\boxed{\sf{tan\theta=\dfrac{Height}{Base}}}

tanθ=

Base

Height

So comparing this formula and value of tan θ from question, we get :

Height = 1

Base = √7

Now we need to Prove the value of : \sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=

4

3

Also :

\large\boxed{\sf{cosec\theta=\dfrac{Hypotenuse}{Height}}}

cosecθ=

Height

Hypotenuse

\large\boxed{\sf{sec\theta=\dfrac{Hypotenuse}{Base}}}

secθ=

Base

Hypotenuse

From this we get :

\large\boxed{\sf{cosec^2\theta=\left(\dfrac{Hypotenuse}{Height}\right)^2}}

cosec

2

θ=(

Height

Hypotenuse

)

2

\large\boxed{\sf{sec^2\theta=\left(\dfrac{Hypotenuse}{Base}\right)^2}}

sec

2

θ=(

Base

Hypotenuse

)

2

But we have Height and Base, we dont have Hypotenuse.

Hypotenuse can be found by using Pythagoras Theorem

Pythagoras Theorem states that :

Hypotenuse² = Side² + Side²

For our question :

Hypotenuse² = Height² + Base²

Hypotenuse² = 1² + √7²

Hypotenuse² = 1 + 7

Hypotenuse² = 8

√Hypotenuse² = √8

Hypotenuse = √8

➢ Let's find value's of cosec²θ and sec²θ

________________________________________

First cosec²θ :

\large\boxed{\sf{cosec^2\theta=\left(\dfrac{Hypotenuse}{Height}\right)^2}}

cosec

2

θ=(

Height

Hypotenuse

)

2

\sf{cosec^2\theta=\left(\dfrac{\sqrt{8}}{1}\right)^2}cosec

2

θ=(

1

8

)

2

\sf{cosec^2\theta=\dfrac{8}{1}}cosec

2

θ=

1

8

cosec²θ = 8

________________________________________

Now sec²θ :

\large\boxed{\sf{sec^2\theta=\left(\dfrac{Hypotenuse}{Base}\right)^2}}

sec

2

θ=(

Base

Hypotenuse

)

2

\sf{sec^2\theta=\left(\dfrac{\sqrt{8}}{\sqrt{7}}\right)^2}sec

2

θ=(

7

8

)

2

\sf{sec^2\theta=\dfrac{8}{7}}sec

2

θ=

7

8

sec²θ = 8/7

________________________________________

Now Proving :

\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=

4

3

Taking L.H.S :

\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }}

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=\sf{\dfrac{8 - sec ^2\theta}{8 + sec^2\theta }}=

8+sec

2

θ

8−sec

2

θ

=\sf{\dfrac{8 - \dfrac{8}{7}}{8 + \dfrac{8}{7} }}=

8+

7

8

8−

7

8

=\sf{\dfrac{\dfrac{48}{7}}{\dfrac{64}{7} }}=

7

64

7

48

\sf{=\dfrac{48\times \:7}{7\times \:64}}=

7×64

48×7

\sf{=\dfrac{48}{64}}=

64

48

\bf{=\dfrac{3}{4}}=

4

3

= R.H.S

Hence Proved !!!

7 0
2 years ago
How Do Solve 3/4=3/8x-3/2 And Show All Your Work
lubasha [3.4K]
It's too hard for me to show all the work, but you can find a common denominator in 8.

Change 3/4 to 6/8 and -3/2 to -12/8.

Now that all the fractions have the same denominator, you can multiply the whole problem by 8 and cancel out all denominators.

You are left with 6=3x-12

Add 12 to 6 and cancel out on the right side.

18=3x

Divide by 18 by 3

x=6
3 0
2 years ago
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