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Lera25 [3.4K]
3 years ago
9

The quotient of a and c is 8

Mathematics
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

1

Step-by-step explanation:

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Why is no one helping or answering ANY of my questions asked so far? :( I thought it said questions are normally answered with-i
vodka [1.7K]
Questions are USUALLY answered within 10 minutes.

Therefore, you're questions might be complicated, hard, or incomplete.

Actual people are answering these questions, so obviously the easier questions get answered first.
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3 years ago
21. In 4 + In(4x - 15) = In(5x + 19)
Alexxx [7]

Answer:

x=-30;\quad \:I\ne \:0

Step-by-step explanation:

In\cdot \:4+In\left(4x-15\right)=In\left(5x+19\right)

\:4+In\left(4x-15\right):\quad -11nI+4nxI\\In\cdot \:4+In\left(4x-15\right)\\=4nI+nI\left(4x-15\right)\\\\\:In\left(4x-15\right):\quad 4nxI-15nI\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac\\a=In,\:b=4x,\:c=15\\=In\cdot \:4x-In\cdot \:15\\=4nxI-15nI\\=In\cdot \:4+4nxI-15nI\\\mathrm{Simplify}\:In\cdot \:4+4nxI-15nI:\quad -11nI+4nxI\\In\cdot \:4+4nxI-15nI\\\mathrm{Group\:like\:terms}\\=4nI-15nI+4nxI\\\mathrm{Add\:similar\:elements:}\:4nI-15nI=-11nI\\=-11nI+4nxI\\

\mathrm{Expand\:}In\left(5x+19\right):\quad 5nxI+19nI\\In\left(5x+19\right)\\=nI\left(5x+19\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=In,\:b=5x,\:c=19\\=In\cdot \:5x+In\cdot \:19\\=5nxI+19nI\\\\-11nI+4nxI=5nxI+19nI\\\\\mathrm{Add\:}11nI\mathrm{\:to\:both\:sides}\\-11nI+4nxI+11nI=5nxI+19nI+11nI\\Simplify\\4nxI=5nxI+30nI\\\mathrm{Subtract\:}5nxI\mathrm{\:from\:both\:sides}\\4nxI-5nxI=5nxI+30nI-5nxI\\\mathrm{Simplify}\\-nxI=30nI\\

\mathrm{Divide\:both\:sides\:by\:}-nI;\quad \:I\ne \:0\\\frac{-nxI}{-nI}=\frac{30nI}{-nI};\quad \:I\ne \:0\\\mathrm{Simplify}\\\frac{-nxI}{-nI}=\frac{30nI}{-nI}\\\mathrm{Simplify\:}\frac{-nxI}{-nI}:\quad x\\\frac{-nxI}{-nI}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{nxI}{nI}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{xI}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=x\\\mathrm{Simplify\:}\frac{30nI}{-nI}:\quad -30\\\mathrm{Apply\:the\:fraction\:rule}:

\quad \frac{a}{-b}=-\frac{a}{b}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{30I}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=-30\\x=-30;\quad \:I\ne \:0

3 0
3 years ago
Midpoint of the segment between the points (−5,13) and (6,4)
Leokris [45]

Answer: (0.5,8.5)

Step-by-step explanation:

We need to use the following formula to find the Midpoint "M":

M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Given the points (-5,13) and (6,4) can identify that:

x_1=-5\\x_2=6\\\\y_1=13\\y_2=4

The final step is to substitute values into the formula.

Therefore, the midpoint of the segment between the points (-5,13) and (6,4) is:

M=(\frac{-5+6}{2},\frac{13+4}{2})\\\\M=(\frac{1}{2},\frac{17}{2})\\\\M=(0.5,8.5)

6 0
3 years ago
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how many meatballs are required fora party of 10 adults and 5 childen? fomual is m=4a+2 where c= children and m=meatball and a=
professor190 [17]
The answer is 42 because there are 10 adults and
M=4(10)+2
M=40+2
M=42

3 0
3 years ago
The area of a triangle is 80x^5y^3. The height of the triangle is x^4y. What is the length of the base of the triangle?
svetoff [14.1K]
Area of a triangle is given by 1/2bh where b is the base and h is the perpendicular height of the triangle.
The area is 80x∧5y³ and the height is x∧4y
Thus; 80x∧5y³ = 1/2(x∧4y) b
         160x∧5y³ = (x∧4y)b
                  b = (160x∧5y³)/ x∧4y)
                  b = 160xy²
Therefore, the base of the triangle is 160xy²
7 0
3 years ago
Read 2 more answers
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