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3 years ago

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The length of 2 sides of a triangle are 12 inches and 4 inches. Which is the third side?

1 answer:

Westkost [7]3 years ago

7 0

Answer:

Let x = the third side

In a triangle, the sum of any 2 sides must be larger than the third side.

I believe this is called the triangle inequality theorem.

We can construct 3 inequalities to obtain the range of values for the third side.

(Inequality #1) 12 + 4 > x

16 > x

(Inequality#2) 12 + x > 4

x > -8 (we can discard this ... we know all sides will be positive)

(Inequality #3) 4 + x > 12

x > 8

So when we combine these together,

8 < x < 16

X (the third side) must be a number between 8 and 16. but not including 8 and 16

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Find the sum of the first 25 terms in this geometric series:<br> 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

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

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

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

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

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

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

$a_1=8$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

so

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

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

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

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

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

$=31.98$

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

3 0

3 years ago

Help me pls math genius

Alecsey [184]

Answer:

x = -3

y = -1

Step-by-step explanation:

in the given question :-

=》x - 4y = 1

=》x = 4y + 1

now replacing the value of x as ( 4y + 1 ) in equation (2)

=》3x + 2y = -11

=》3 (4y + 1) + 2y = -11

=》12y + 3 + 2y = -11

=》14y = -11 - 3

=》y = -14 ÷ 14

=》y = -1

now, putting the value of y in equation (1)

=》x - 4y = 1

=》x - ( 4 × -1 ) = 1

=》x + 4 = 1

=》x = 1 - 4

=》x = -3

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Semenov [28]

The denominator would still be the same you placed the marble back into the bag

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3 years ago

Find the product -2X²×4X²y²

nydimaria [60]

Answer:

-8X^4y^2

Step-by-step explanation:

-2X² × 4X²y² = -8X^4y^2

edit to add:

first: multiply the coefficients together:

-2 x 4 = -8

second: add the exponents of X together (when you multiply exponents, you just add them together):

² & ² = ^4

and you have the answer:

-8X^4y^2

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