CAN SOMEONE HELP ON THESE QUESTIONS PLEASE!!!!

Mathematics

1 answer:

BaLLatris [955]3 years ago

6 0

Answer:

use photomath

Step-by-step explanation:

it works

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Ill give brainliest - please help ASAP WITH THE WORK THO PLEASE?

ohaa [14]

Answer:

we have.

11x=½(16x+16+50)

22x=16x+66

22x-16x=66

6x=66

x=66/6

x=11

118°=½(6x-11+181)

236=6x+170

6x=236-170

x=66/6

x=11°

34°=½(arcAK-arc LN)

34×2=(110-arc LN)

arc LN =110-68

arc LN=42°

<L=½(arc EN-arc KM)

<L=½(139°-73°)

<L=33

4 0

3 years ago

Solve 12^x^2+5x-4 = 12^2x+6

faust18 [17]

The solutions for ‘x’ are 2 and -5

Step-by-step explanation:

Given equation:

$12^{x^{2}+5 x-4}=12^{2 x+6}$

Since the base on both sides as ‘12’ are the same, we can write it as

$x^{2}+5 x-4=2 x+6$

$x^{2}+5 x-2 x-4-6=0$

$x^{2}+3 x-10=0$

Often, the value of x is easiest to solve by $a x^{2}+b x+c=0$ by factoring a square factor, setting each factor to zero, and then isolating each factor. Whereas sometimes the equation is too awkward or doesn't matter at all, or you just don't feel like factoring.