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AleksandrR [38]
3 years ago
5

rac{1}{12} + \frac{6}{12} " alt=" \frac{1}{12} + \frac{6}{12} " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
marin [14]3 years ago
3 0

1/12+6/12

=1+6/12

=7/12

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Which expression is equivalent to ?
mamaluj [8]

Answer:

Step-by-step explanation:

\frac{(a^{2}b^{4}c)^{2}*(6a^{3}b)*(2c^{5})^{3}}{4a^{6}b^{12}c{^3}}\\\\=\frac{a^{2*2}b^{4*2}c^{2}6a^{3}b2^{3}c^{5*3}}{4a^{6}b^{12}c{^3}}\\\\=\frac{a^{4}b^{8}c^{2}6a^{3}b8c^{15}}{4a^{6}b^{12}c{^3}}\\\\=\frac{6*8*a^{4+3}b^{8+1}c^{2+15}}{4a^{6}b^{12}c{^3}}\\\\=\frac{6*2*a^{7}b^{9}c^{17}}{a^{6}b^{12}c{^3}}\\\\=\frac{12a^{7-6}c{17-3}}{b^{12-9}}\\\\=\frac{12a^{1}c^{14}}{b^{3}}\\\\=\frac{12ac^{14}}{b^{3}}

Hint:

x^{m}*x^{n}=x^{m+n}\\\frac{x^{m}}{x^{n}}=x^{m-n},m>n\\\frac{x^{m}}{x^{n}}=\frac{1}{x^{n-m}},n>m

5 0
3 years ago
Kyle lives 3/4 mile from his friend Aiden. Kyle is to walk 2/3 of the distance to meet Aiden. What part of a mile must Kyle walk
bearhunter [10]

Answer:

1/2 mile

Step-by-step explanation:

\frac{2}{3} (\frac{3}{4} )=\frac{6}{12} =\frac{1}{2}

6 0
3 years ago
The electrician charges $50 per visit plus $12 per hour of work. Steve hires the electrician to do some work for his house. If S
TEA [102]

Answer:

12.5 hours

Step-by-step explanation:

budget: 200

fix cost: 50

hourly rate: 12

hours: ?

----

First we deduct the fix 50 dollars cost from the total budget.

Then we divide the remaining amount with the hourly rate

( 200 - 50 ) / 12 = ?

? = 12.5 hours

8 0
3 years ago
The measure of an angle is 108°. What is the measure of a supplementary angle?
gogolik [260]

Answer:

72 degrees

Step-by-step explanation:

180-108=72

where I got 180: a supplementary angle is 180 degrees

8 0
3 years ago
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.
konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

BC^2=BD\cdot AB.

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.

Take positive value x. You get

AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.

2. According to the previous theorem,

AC^2=AD\cdot AB.

Then

AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

Answer: AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D

This means that you cannot find solutions of this equation. Then CD≠2 cm.

8 0
3 years ago
Read 2 more answers
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