Find the remainder when $3 \times 13 \times 23 \times 33 \times \ldots \times 183 \times 193$ is divided by $5$.

Mathematics

1 answer:

Trava [24]3 years ago

5 0

209.095.543and 8 tenths

You might be interested in

Share £747 in the ratio 2:7 between Tom and Ben. Tom will get £ Ben will get £

kari74 [83]

$tom \: will \: get \:£166 \\ ben \: will \: get \: £581\\ solution \\ let \: the \: ratios \: be \: 2x \: and \: 7x \\ 2x + 7x = 747 \\ or \: 9x = 747 \\ or \: x = \frac{747}{9} \\ x = £83 \\ replacing \: value \\ 2x \\ = 2 \times 83 \\ = 166 \\ 7x \\ = 7 \times 83 \\ = 581 \\ hope \: it \: helps$

5 0

3 years ago

Solve for x. Express the answer in simplest form.1/7 = 3/4x - 1

shtirl [24]

1/7 = 3/4x-1
3/4x=1/7+1=8/7
x=8/7*4/3=32/21

4 0

3 years ago

Find an equation of the plane consisting of all points that are equidistant from (-3, 5, -4) and (-5, 0, 4), and having -2 as th

Firdavs [7]

Answer:

-2x-5y+8z+4.5=0

Step-by-step explanation:

Let (x,y,z) be the coordinates of the point lying on the needed plane. This point is equidistant from the points (-3, 5, -4) and (-5, 0, 4), so

$d_1=\sqrt{(x-(-3))^2+(y-5)^2+(z-(-4))^2}=\sqrt{(x+3)^2+(y-5)^2+(z+4)^2}\\ \\d_2=\sqrt{(x-(-5))^2+(y-0)^2+(z-4)^2}=\sqrt{(x+5)^2+y^2+(z-4)^2}\\ \\d_1=d_2\Rightarrow \sqrt{(x+3)^2+(y-5)^2+(z+4)^2}=\sqrt{(x+5)^2+y^2+(z-4)^2}\\ \\(x+3)^2+(y-5)^2+(z+4)^2=(x+5)^2+y^2+(z-4)^2\\ \\x^2+6x+9+y^2-10y+25+z^2+8z+16=x^2+10x+25+y^2+z^2-8z+16\\ \\-4x-10y+16z+9=0\\ \\-2x-5y+8z+4.5=0$