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3 years ago

Pl0x halp fast DDDD:

Mathematics

1 answer:

N76 [4]3 years ago

8 0

(Lets hope i'm right.)

1. A I think.

2. A since its a question

Have a lovely day! ~Pooch ♥

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Simplify the expression:<br> -6x-4x-10+3

tangare [24]

Answer:

-10x-7

Step-by-step explanation:

6 0

3 years ago

Let a, b, c and x elements in the group G. In each of the following solve for x in terms of a, b, and c.

alina1380 [7]

Answer:

The answer is $x=a^{-1}cb^{-1}$ .

Step-by-step explanation:

First, it is important to recall that the group law is not commutative in general, so we cannot assume it here. In order to solve the exercise we need to remember the axioms of group, specially the existence of the inverse element, i.e., for each element $g\in G$ there exist another element, denoted by $g^{-1}$ such that $gg^{-1}=e$ , where $e$ stands for the identity element of G.

So, given the equality $axb=c$ we make a left multiplication by $a^{-1}$ and we obtain:

$a^{-1}axb =a^{-1}c.$

But, $a^{-1}axb = exb = xb$ . Hence, $xb = a^{-1}c$ .

Now, in the equality $xb = a^{-1}c$ we make a right multiplication by $b^{-1}$ , and we obtain

$xbb^{-1} = a^{-1}cb^{-1}$ .

Recall that $bb^{-1}=e$ and $xe=x$ . Therefore,

$x=a^{-1}cb^{-1}$ .

6 0

2 years ago

PLEASE PLEASE HELP ME OUT! But i ask that you explain how you got your answer and not just give the answer.

rjkz [21]

ok so first we would do the function d(x) = 3(8)+2

which when solved turns out to be 26

we would then do the function h(x) = 2(26) +7

which when solved turns out to be 59

so he would need 59 dollars for the donuts

6 0

3 years ago

Find 2 common angles that sum to (17pi/12) 2. Evaluate tan(17pi/12) using the sum identity for tangent.

Dmitry [639]

Note that
$\frac{17 \pi }{12} = \frac{3 \pi }{12} + \frac{14 \pi }{12} = \frac{ \pi }{4} + \frac{7 \pi }{6}$

Note that
$x= \frac{ \pi }{4}:\,\, sin(x) =cos(x)= \frac{1}{ \sqrt{2} },\,tan(x)=1\\x= \frac{7 \pi }{6} :\,\,sin(x)=- \frac{1}{2} ,\,\,cos(x)=- \frac{ \sqrt{3} }{2} ,\,\,tan(x)= \frac{1}{ \sqrt{3} }$

Use the identity
$tan(x+y)= \frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

Therefore
$tan( \frac{17 \pi }{12} )= \frac{1+ \frac{1}{ \sqrt{3} } }{1- \frac{1}{ \sqrt{3} } } = \frac{ \sqrt{3}+1 }{ \sqrt{3}-1} = \frac{( \sqrt{3}+1 )^{2}}{( \sqrt{3}-1 )( \sqrt{3}+1 )} = \frac{3+1+2 \sqrt{3}}{3-1} =2+ \sqrt{3}$

Answer: 2 + √3

8 0

2 years ago

Solve for m.t=mr^2/3

spin [16.1K]

t = mr^2/3

Divide both sides by r^2/3

m = t / r^2/3

6 0

2 years ago

Read 2 more answers