All categories

3 years ago

Find the product (r-9)(r+4)

Mathematics

2 answers:

AVprozaik [17]3 years ago

7 0

Answer:

2r - 36

Step-by-step explanation:

r x r = 2r

-9 x 4 = -36

Mrrafil [7]3 years ago

3 0

Answer:

$r^{2} -5r-36$

You might be interested in

If you got 5 dogs and 7 cats but one got eaten from both of them. How much do you have left.

Talja [164]

4 & 6 cause one of both of them got eaten

4 0

3 years ago

If a/b and c/d are rational expressions, then a/b•c/d=a•c/b•d <br><br> true or false?

alukav5142 [94]

That would be true :)

5 0

3 years ago

Need help. I can’t fail this!! 3

Ksju [112]

Answer:

it's A

Step-by-step explanation:

6 0

3 years ago

Bob is buying the soccer team ice cream after the game today eight players my ice cream cookies in five please my ice cream cone

blsea [12.9K]

Answer:$18.75

Step-by-step explanation: first multiply 8 times 1.25 then multiply 5 times 1.75 then add both of your numbers and put the decimal point back

5 0

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