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2 years ago

What is the next number? 2, 7, 26, 101, 400, ?

Mathematics

2 answers:

Y_Kistochka [10]2 years ago

8 0

7/2=3.5
26/7=3.7
101/26=3.9
400/101=4
400 x 4 or 4.1 = 1600 or 1640

densk [106]2 years ago

3 0

It is 1595. Put it on excel so you can better have an overview how it is done. You need 5 columns, column A-E. On column A, put 1 to 5 in rows 1 to 5. On column C, put the answers: 2, 7, 26, 101 and 400. On column D, put number 3 in all 5 rows. Back on column B, insert =sum(C1-A1) on row 1 and copy till row 5. On column E, insert =sum(B1+C1*D1) on row 1 and copy till row 5 and you will get 1595 at the end.

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Hi,so I've tried solving this problem for a while and I just don't get it,if you could give me some ideas I would appreciate it.

ziro4ka [17]

Answer:

187 cm²

Step-by-step explanation:

The bottom rectangle area is easy, it is 15*9 = 135 cm².

To find the area of the triangle, you only need the base width and its height (you don't need the hypotenuse). You can then use the formula: area triangle is base times half height.

The base width is 15-7 = 8 cm

The height is 22-9 = 13 cm

So the area of the triangle is 13*8/2 = 52 cm²

Together with the 135 of the rectangle that sums to 52+135 = 187 cm².

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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:

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

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

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

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

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

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3 years ago

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Yakvenalex [24]

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2 years ago

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