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3 years ago

7

For what values (cases) of the variables the expression does not exist:

2%7D" id="TexFormula1" title="\frac{3}{4y+2}" alt="\frac{3}{4y+2}" align="absmiddle" class="latex-formula">

1 answer:

stealth61 [152]3 years ago

6 0

Answer:

y = - $\frac{1}{2}$

Step-by-step explanation:

Given

$\frac{3}{4y+2}$

The denominator of the expression cannot be zero as this would make it undefined. Equating the denominator to zero and solving gives the value that y cannot be, that is

4y + 2 = 0 ( subtract 2 from both sides )

4y = - 2 ( divide both sides by 4 )

y = $\frac{-2}{4}$ = - $\frac{1}{2}$ ← excluded value

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The sum of 3 odd consecutive integers are 93. find the 3 integers

Solnce55 [7]

Let, Your Integers = x, x+2, x+4
Now, x + x+2 + x+4 = 93
3x + 6 = 93
3x = 93 - 6
x = 87/3
x = 29
Then, x+2 = 31, & x+4 = 33

In short, Your Integers would be 29, 31, 33

Hope this helps!

4 0

3 years ago

Please dont ignore, Need help!!! Use the law of sines/cosines to find..

Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

<h3>16</h3>

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

$\sin{A} = \sin{103\textdegree{}}$ ,
The opposite side of angle A $a = BC = 26$ ,
The angle C is to be found, and
The length of the side opposite to angle C $c = AB = 6$ .

$\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}$ .

$\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}$ .

$\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}$ .

Note that the inverse sine function here $\sin^{-1}()$ is also known as arcsin.

<h3>17</h3>

By the law of cosine,

$c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C}$ ,

where

$a$ , $b$ , and $c$ are the lengths of sides of triangle ABC, and
$\cos{C}$ is the cosine of angle C.

For triangle ABC:

$b = 21$ ,
$c = 30$ ,
The length of $a$ (segment BC) is to be found, and
The cosine of angle A is $\cos{123\textdegree}$ .

Therefore, replace C in the equation with A, and the law of cosine will become:

$a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}$ .

$\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}$ .

<h3>18</h3>

For triangle ABC:

$a = 14$ ,
$b = 9$ ,
$c = 6$ , and
Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

$b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}$ .

$\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}$ .

$\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree$ .

<h3>15</h3>

For triangle DEF:

The length of segment DF is to be found,
The length of segment EF is 9,
The sine of angle E is $\sin{64\textdegree}}$ , and
The sine of angle D is $\sin{39\textdegree}$ .

Apply the law of sine:

$\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}$

$\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9$ .

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3 years ago

F (x) = v= - 8. Find f '(x) and its domain.

Flauer [41]

Answer:

F

Step-by-step explanation:

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2 years ago

Rewrite 56+32 as the product of the GCF and a sum.

Vitek1552 [10]

56 = 2 x 2 x 2 x 732 = 2 x 2 x 2 x 2 x 2
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3 years ago

Give me an example of an irrational number that is greater than 10

Pavel [41]

To find:

An irrational number that is greater than 10.

Solution:

Irritation number: It cannot be expression in the form of $\dfrac{p}{q}$ , where, $q\neq 0, p,q$ are integers.

For example: $\sqrt{2}\sqrt{3},\pi, 1.263689...,etc.$ .

We know that square of 10 is 100. So, square root of any prime number is an example of an irrational number that is greater than 10.

First prime number after 100 is 101.

Required irrational number $=\sqrt{101}$

Therefore, $\sqrt{101}$ is an irrational number that is greater than 10.

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3 years ago