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2 years ago

Two lines intersecting at a right angle ?

Mathematics

2 answers:

MAXImum [283]2 years ago

5 0

Answer:

are perpendicular

Citrus2011 [14]2 years ago

5 0

Yes, it is perpendicular

Parallel is when you have two lines that do not intersect

A ray is a line that has one definite starting point and no obvious end point. (Stars with a point and "ends" with an arrow, meaning that it goes on "forever")

A line is just one specific line

When you have perpendicular, that is when two lines cross each other forming a right angle as you may now know

Hope this helps :)

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Write the standard form of the line that passes through the given points (6,1) and (5,4)

Serggg [28]

The standard form:

$Ax+By=C$

The point-slope form:

$y-y_1=m(x-x_1)$

m - slope

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (6, 1) and (5, 4). Substitute:

$m=\dfrac{4-1}{5-6}=\dfrac{3}{-1}=-3$

$y-1=-3(x-1)$ use distributive property

$y-1=-3x+3$ add 1 to both sides

$y=-3x+4$ add 3x to both sides

$\boxed{3x+y=4}$

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3 years ago

Find the sum of the first 25 terms in this geometric series: 8 + 6 + 4.5...

Ksivusya [100]

Step-by-step explanation:

Given the geometric sequence

8 + 6 + 4.5...

A geometric sequence has a constant ratio and is defined by

$a_n=a_1\cdot r^{n-1}$

$\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\quad \:r=\frac{a_{n+1}}{a_n}$

$\frac{6}{8}=\frac{3}{4},\:\quad \frac{4.5}{6}=\frac{3}{4}$

$\mathrm{The\:ratio\:of\:all\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}$

$r=\frac{3}{4}$

$\mathrm{The\:first\:element\:of\:the\:sequence\:is}$

$a_1=8$

$\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:$

$a_n=8\left(\frac{3}{4}\right)^{n-1}$

$\mathrm{Geometric\:sequence\:sum\:formula:}$

$a_1\frac{1-r^n}{1-r}$

$\mathrm{Plug\:in\:the\:values:}$

$n=25,\:\spacea_1=8,\:\spacer=\frac{3}{4}$

$=8\cdot \frac{1-\left(\frac{3}{4}\right)^{25}}{1-\frac{3}{4}}$

$\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}$

$=\frac{\left(1-\left(\frac{3}{4}\right)^{25}\right)\cdot \:8}{1-\frac{3}{4}}$

$=\frac{8\left(-\left(\frac{3}{4}\right)^{25}+1\right)}{\frac{1}{4}}$

$\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}$

$=\frac{8\left(-\frac{3^{25}}{4^{25}}+1\right)}{\frac{1}{4}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{\frac{b}{c}}=\frac{a\cdot \:c}{b}$

$=\frac{\left(1-\frac{3^{25}}{4^{25}}\right)\cdot \:8\cdot \:4}{1}$

$\mathrm{Multiply\:the\:numbers:}\:8\cdot \:4=32$

$=\frac{32\left(-\frac{3^{25}}{4^{25}}+1\right)}{1}$

$=\frac{32\cdot \frac{4^{25}-3^{25}}{4^{25}}}{1}$ ∵ $\mathrm{Join}\:1-\frac{3^{25}}{4^{25}}:\quad \frac{4^{25}-3^{25}}{4^{25}}$

$=32\cdot \frac{4^{25}-3^{25}}{4^{25}}$

$=\frac{\left(4^{25}-3^{25}\right)\cdot \:32}{4^{25}}$

$=\frac{2^5\left(4^{25}-3^{25}\right)}{2^{50}}$ ∵ $\mathrm{Factor}\:32:\ 2^5$ , $\mathrm{Factor}\:4^{25}:\ 2^{50}$

$=\frac{4^{25}-3^{25}}{2^{45}}$ ∵ $\mathrm{Cancel\:}\frac{\left(4^{25}-3^{25}\right)\cdot \:2^5}{2^{50}}:\quad \frac{4^{25}-3^{25}}{2^{45}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a\pm \:b}{c}=\frac{a}{c}\pm \frac{b}{c}$

$=\frac{4^{25}}{2^{45}}-\frac{3^{25}}{2^{45}}$

$=32-\frac{3^{25}}{2^{45}}$ ∵ $\frac{4^{25}}{2^{45}}=32$

$=32-0.024$ ∵ $\frac{3^{25}}{2^{45}}=0.024$

$=31.98$

Therefore, the sum of the first 25 terms in this geometric series: 31.98

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Answer:

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