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Troyanec [42]
3 years ago
12

Two problems here I need solved! I need every step, so please have that with your answers!!

Mathematics
1 answer:
Free_Kalibri [48]3 years ago
7 0
QUESTION 1

We want to solve,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}

We factor the denominator of the fraction on the right hand side to get,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.

This implies
\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}

We multiply through by LCM of
(x-4)(x - 2)

(x - 2) + x(x-4) = 2

We expand to get,

x - 2 + {x}^{2} - 4x= 2

We group like terms and equate everything to zero,

{x}^{2} + x - 4x - 2 - 2 = 0

We split the middle term,

{x}^{2} + - 3x - 4 = 0

We factor to get,

{x}^{2} + x - 4x- 4 = 0

x(x + 1) - 4(x + 1) = 0

(x + 1)(x - 4) = 0

x + 1 = 0 \: or \: x - 4 = 0

x = - 1 \: or \: x = 4

But
x = 4
is not in the domain of the given equation.

It is an extraneous solution.

\therefore \: x = - 1
is the only solution.

QUESTION 2

\sqrt{x+11} -x=-1

We add x to both sides,

\sqrt{x+11} =x-1

We square both sides,

x + 11 = (x - 1)^{2}

We expand to get,

x + 11 = {x}^{2} - 2x + 1

This implies,

{x}^{2} - 3x - 10 = 0

We solve this quadratic equation by factorization,

{x}^{2} - 5x + 2x - 10 = 0

x(x - 5) + 2(x - 5) = 0

(x + 2)(x - 5) = 0

x + 2 = 0 \: or \: x - 5 = 0

x = - 2 \: or \: x = 5

But
x = - 2
is an extraneous solution

\therefore \: x = 5
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Step-by-step explanation:

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Please solve this, will rate 5 stars and mark as STAR!​
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Answer:

\boxed{5 \cdot \sqrt{2}  \cdot \sqrt[6]{5} }

Step-by-step explanation:

\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }

\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}

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\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }

250=2 \cdot 5^3

\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5  \sqrt[3]{2}

Once

\sqrt[6]{2}  \cdot \sqrt[6]{5} = \sqrt[6]{10}

We have

5  \sqrt[3]{2} \cdot \sqrt[6]{2}  \cdot \sqrt[6]{5}

We can proceed considering the common base of exponentials

\sqrt[3]{2}  \cdot \sqrt[6]{2}  =  2^{\frac{1}{3}} \cdot  2^{\frac{1}{6} }  = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}

Therefore,

5  \sqrt[3]{2} \cdot \sqrt[6]{2}  \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2}  \cdot \sqrt[6]{5}

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3 years ago
A tree that is 14 feet tall casts a shadow that is 30 feet long. Find the angle of the elevation from the top of the shadow to t
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Para decorar una pared se disponen de tiras de papel azules de 15 cm, verdes de 20 cm, y rojas de 25 cm. En la pared se quiere a
den301095 [7]

Answer:

a) Smallest line that can be made with each color = 300 cm

b) Total strips should be used = 47 strips

c) Total strips used of blue color = 20

  Total strips used of green color = 15

  Total strips used of red color = 12

Step-by-step explanation:

Given - To decorate a wall, there are 15 cm blue, 20 cm green, and 25 cm red strips of paper. On the wall you want to build three lines of the same size, one of each color and without cutting any strip.

To find - a) How long is the smallest line that can be made with each color?

              b) How many strips should be used?

              c) How many of each color?

Proof -

a)

For the smallest line that can be made with each color, we just have to find the lcm (least common multiple) of the 3 srtips.

Firstly,

Decompose the 3 strips to its prime factors , we get

15 = 3×5

20 = 2²×5

25 = 5²

So,

The Lcm(15, 20, 25) = 3×2²×5² = 3×4×25 = 300

∴ we get

Smallest line that can be made with each color = 300 cm

b)

Now,

Total strips used = 300 cm

Strips used by 15 cm blue = \frac{300}{15} = 20 strips

Strips used by 20 cm green = \frac{300}{20} = 15 strips

Strips used by 25 cm red = \frac{300}{25} = 12 strips

So,

Total strips should be used = 20 + 15 + 12 = 47 strips

c)

Total strips used of blue color = 20

Total strips used of green color = 15

Total strips used of red color = 12

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2 years ago
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Answer:

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An absolute value has 2 solutions, a positive and a negative.  Remember to flip the inequality for the negative solution

|5x-4|>16

5x-4 > 16        5x-4 < -16

Add 4 to each side

5x-4 +4> 16+4   or     5x-4+4 < -16+4

5x > 20                or        5x < -12

Divide each side by 5

5x/5 > 20/5          or         5x/5 < -12/5

x >4                         or            x < -12/5

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