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sertanlavr [38]
3 years ago
8

Please help me solve this problem I’m really stuck on!!! Please help right now!!!

Mathematics
1 answer:
valentinak56 [21]3 years ago
5 0

\frac{ ({x}^{1}  {y}^{ \frac{1}{4} })^{2}  }{ {x}^{ \frac{5}{4} }  {y}^{ \frac{5}{8} } } \\   \frac{ {x}^{2} {y}^{ \frac{2}{4} }  }{ {x}^{ \frac{5}{4} } {y}^{ \frac{5}{8} }  }  \\   \frac{ {x}^{2}  {y}^{ \frac{1}{2} } }{ {x}^{ \frac{5}{4} }  {y}^{ \frac{5}{8} } }  \\

You can distribute the exponent outside of the parentheses by multiplying it to the exponent in each term in the numerator

When dividing, subtract the exponent of the numerator by the exponent of the denominator for each variable

For x:

{x}^{2}  -  {x}^{ \frac{5}{4} }  \\  {x}^{ \frac{3}{4} }

For y:

{y}^{ \frac{1}{2} }  -  {y}^{ \frac{5}{8} }  \\  {y}^{ \frac{ - 1}{8} }  \\  \sqrt[8]{y}

EDIT: y^(-1/8) is equal to 1 / (8 radical y)

Final answer:

\frac{ {x}^{ \frac{3}{4} } }{ \sqrt[8]{y} }

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Use a flowchart to prove if the triangles in each pair are congruent. NO LINKS!!!!​
valentinak56 [21]

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Answer:

  see attached

Step-by-step explanation:

It isn't clear what is supposed to go in the various blanks. We have elected to identify the corresponding congruent parts, and name the congruent triangles. The postulate supporting the conclusion is also shown.

In most cases, corresponding parts are marked congruent. The exception is the vertical angles in figure 22.

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. (02.05 MC) Side AB = 5, side BC = 6, side DE = 5, and side EF = 6. What additional information would you need to prove that ΔA
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Answer:

Side AC is congruent to side DF

Step-by-step explanation:

We are given that

Side AB=5 cm

BC=6 cm

Side DE=5 cm

Side EF= 6cm

We have to find an additional information would  need to prove that \triangle ABC\cong \triangle DEF by SSS.

When two triangles are congruent by SSS it means three sides of one triangle are congruent to corresponding sides of other triangle.

We have in triangle ABC and triangle DEF

AB\cong DE=5 cm

BC\cong EF=6 cm

AC\cong DF

Then ,\triangle ABC\cong \triangleDEF

Reason: SSS postulates

Answer: Side AC is congruent to side DF

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ANSWER ASAP (17 points)
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Answer:

A.) x > 50

Step-by-step explanation:

hopefully my answer helps! ♡

(if my answers helped you please tell me, therefore I know.. that I did them correctly or incorrectly. anyways, have a great day/night!)

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A right triangle has an area of 18 The length of each leg is 6 inches. If the triangle is not an isosceles triangle, what are al
ladessa [460]

Given:

Area of a right triangle = 18 sq. inches

Length of one leg = 6 inches

To find:

The length of another leg.

Solution:

We know that, the area of a triangle is

A=\dfrac{1}{2}\times base\times height

In a right angle triangle, the area of the triangle is

A=\dfrac{1}{2}\times leg_1\times leg_2

Putting A=18 and leg_1=6, we get

18=\dfrac{1}{2}\times 6\times leg_2

18=3\times leg_2

Divide both sides by 3.

\dfrac{18}{3}=leg_2

6=leg_2

Both legs are equal so the given triangle is an isosceles right triangle.

If the triangle is not isosceles triangle, then the length of legs are factors of 36 because half of 36 is 18.

Therefore, the possible pairs of legs are 1 and 36, 2 and 18, 3 and 12, 4 and 9, 6 and 6, 9 and 4, 12 and 3, 18 and 2, 36 and 1.

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3 years ago
Complete the steps to solve the polynomial equation x3 – 21x = –20. According to the rational root theorem, which number is a po
jeka94

Answer:

Zeroes : 1, 4 and -5.

Potential roots: \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20.

Step-by-step explanation:

The given equation is

x^3-21x=-20

It can be written as

x^3+0x^2-21x+20=0

Splitting the middle terms, we get

x^3-x^2+x^2-x-20x+20=0

x^2(x-1)+x(x-1)-20(x-1)=0

(x-1)(x^2+x-20)=0

Splitting the middle terms, we get

(x-1)(x^2+5x-4x-20)=0

(x-1)(x(x+5)-4(x+5))=0

(x-1)(x+5)(x-4)=0

Using zero product property, we get

x-1=0\Rightarrow x=1

x-4=0\Rightarrow x=4

x+5=0\Rightarrow x=-5

Therefore, the zeroes of the equation are 1, 4 and -5.

According to rational root theorem, the potential root of the polynomial are

x=\dfrac{\text{Factor of constant}}{\text{Factor of leading coefficient}}

Constant = 20

Factors of constant ±1, ±2, ±4, ±5, ±10, ±20.

Leading coefficient= 1

Factors of leading coefficient ±1.

Therefore, the potential root of the polynomial are \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20.

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