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horsena [70]
3 years ago
7

Can you show the work how how to do it​

Mathematics
2 answers:
Fofino [41]3 years ago
6 0

Answer:uhhh

Step-by-step explanation:

You didn’t include a picture or any real problem

Soloha48 [4]3 years ago
4 0

Answer: we need more information

Step-by-step explanation:

You might be interested in
Simplify completely <br> (X^2+x-12/x^2-x-20)/(3x^2–24x+45/12x^2-48-60)
Rudik [331]

Answer:

(4 (x^4 - 20 x^2 - 12))/(3 x^2 (9 x^2 - 32 x - 144))

Step-by-step explanation:

Simplify the following:

(x^2 + x - x - 20 - 12/x^2)/((15 x^2)/4 + 3 x^2 - 24 x - 60 - 48)

Hint: | Put the fractions in x^2 + x - x - 20 - 12/x^2 over a common denominator.

Put each term in x^2 + x - x - 20 - 12/x^2 over the common denominator x^2: x^2 + x - x - 20 - 12/x^2 = x^4/x^2 + x^3/x^2 - x^3/x^2 - (20 x^2)/x^2 - 12/x^2:

(x^4/x^2 + x^3/x^2 - x^3/x^2 - (20 x^2)/x^2 - 12/x^2)/((45 x^2)/12 + 3 x^2 - 24 x - 60 - 48)

Hint: | Combine x^4/x^2 + x^3/x^2 - x^3/x^2 - (20 x^2)/x^2 - 12/x^2 into a single fraction.

x^4/x^2 + x^3/x^2 - x^3/x^2 - (20 x^2)/x^2 - 12/x^2 = (x^4 + x^3 - x^3 - 20 x^2 - 12)/x^2:

((x^4 + x^3 - x^3 - 20 x^2 - 12)/x^2)/((45 x^2)/12 + 3 x^2 - 24 x - 60 - 48)

Hint: | Group like terms in x^4 + x^3 - x^3 - 20 x^2 - 12.

Grouping like terms, x^4 + x^3 - x^3 - 20 x^2 - 12 = x^4 - 20 x^2 - 12 + (x^3 - x^3):

(x^4 - 20 x^2 - 12 + (x^3 - x^3))/(x^2 ((45 x^2)/12 + 3 x^2 - 24 x - 60 - 48))

Hint: | Look for the difference of two identical terms.

x^3 - x^3 = 0:

((x^4 - 20 x^2 - 12)/x^2)/((45 x^2)/12 + 3 x^2 - 24 x - 60 - 48)

Hint: | In (45 x^2)/12, the numbers 45 in the numerator and 12 in the denominator have gcd greater than one.

The gcd of 45 and 12 is 3, so (45 x^2)/12 = ((3×15) x^2)/(3×4) = 3/3×(15 x^2)/4 = (15 x^2)/4:

(x^4 - 20 x^2 - 12)/(x^2 (15 x^2/4 + 3 x^2 - 24 x - 60 - 48))

Hint: | Put the fractions in (15 x^2)/4 + 3 x^2 - 24 x - 60 - 48 over a common denominator.

Put each term in (15 x^2)/4 + 3 x^2 - 24 x - 60 - 48 over the common denominator 4: (15 x^2)/4 + 3 x^2 - 24 x - 60 - 48 = (15 x^2)/4 + (12 x^2)/4 - (96 x)/4 - 240/4 - 192/4:

(x^4 - 20 x^2 - 12)/(x^2 (15 x^2)/4 + (12 x^2)/4 - (96 x)/4 - 240/4 - 192/4)

Hint: | Combine (15 x^2)/4 + (12 x^2)/4 - (96 x)/4 - 240/4 - 192/4 into a single fraction.

(15 x^2)/4 + (12 x^2)/4 - (96 x)/4 - 240/4 - 192/4 = (15 x^2 + 12 x^2 - 96 x - 240 - 192)/4:

(x^4 - 20 x^2 - 12)/(x^2 (15 x^2 + 12 x^2 - 96 x - 240 - 192)/4)

Hint: | Group like terms in 15 x^2 + 12 x^2 - 96 x - 240 - 192.

Grouping like terms, 15 x^2 + 12 x^2 - 96 x - 240 - 192 = (12 x^2 + 15 x^2) - 96 x + (-192 - 240):

(x^4 - 20 x^2 - 12)/(x^2 ((12 x^2 + 15 x^2) - 96 x + (-192 - 240))/4)

Hint: | Add like terms in 12 x^2 + 15 x^2.

12 x^2 + 15 x^2 = 27 x^2:

(x^4 - 20 x^2 - 12)/(x^2 (27 x^2 - 96 x + (-192 - 240))/4)

Hint: | Evaluate -192 - 240.

-192 - 240 = -432:

(x^4 - 20 x^2 - 12)/(x^2 (27 x^2 - 96 x + -432)/4)

Hint: | Factor out the greatest common divisor of the coefficients of 27 x^2 - 96 x - 432.

Factor 3 out of 27 x^2 - 96 x - 432:

(x^4 - 20 x^2 - 12)/(x^2 (3 (9 x^2 - 32 x - 144))/4)

Hint: | Write ((x^4 - 20 x^2 - 12)/x^2)/((3 (9 x^2 - 32 x - 144))/4) as a single fraction.

Multiply the numerator by the reciprocal of the denominator, ((x^4 - 20 x^2 - 12)/x^2)/((3 (9 x^2 - 32 x - 144))/4) = (x^4 - 20 x^2 - 12)/x^2×4/(3 (9 x^2 - 32 x - 144)):

Answer: (4 (x^4 - 20 x^2 - 12))/(3 x^2 (9 x^2 - 32 x - 144))

3 0
3 years ago
Which trigonometric ratios are correct for triangle ABC? Check all that apply
sergij07 [2.7K]
Consider an angle M with measure m≠90°, in a right triangle.

Let

OPP denote the length of the side opposite to M,
ADJ denote the length of the side adjacent to M, and 
HYP denote the hypotenuse.

then: 

Sin(M) = OPP/HYP
Cos(M)= ADJ/HYPP
Tan(M)=OPP/ADJ


Back to our problem, 

using the Pythagorean we can find the length of AB: 

|AB|^2+|AC|^2=|BC|^2\\\\|AB|^2+9^2=18^2\\\\|AB|^2=18^2-9^2\\\\|AB|^2=(18-9)(18+9)=9 \cdot 27=9 \cdot 9 \cdot3\\\\|AB|=9 \sqrt{3} &#10;

Sin(C)= \frac{OPP}{HYP}=\frac{9 \sqrt{3} }{18}= \frac{ \sqrt{3} }{2}
Cos(B)= \frac{ADJ}{HYP}= \frac{9 \sqrt{3}}{18}= \frac{ \sqrt{3}}{2}
Tan(C)= \frac{OPP}{ADJ}= \frac{9 \sqrt{3} }{9}= \sqrt{3}
Sin(B)= \frac{OPP}{HYP}= \frac{9}{18}= \frac{1}{2}
Tan(B)= \frac{OPP}{ADJ}= \frac{9}{9 \sqrt{3}}= \frac{1}{ \sqrt{3} }= \frac{ \sqrt{3}}{3}


Answer: 1, 3, 4



5 0
3 years ago
Please important answer, as soon as possible I will thank you and mark as brainlist
denis23 [38]
10:12
15:18
i got this from doubling the first to make 5 = 10 and then that made 6 = 12. the second one was the same but tripled to make 6 = 18 and then that made 5 = 15
3 0
4 years ago
Determine which letter best shows the location of the fraction.
Crank
Is there an imagine you forgot to attach to the question? would love to help but i don’t see any options :(
7 0
3 years ago
Who wanna give me the answer ;)
patriot [66]

Answer:

D

Step-by-step explanation:

6 0
3 years ago
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