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marta [7]
2 years ago
5

Trigonometry help; how would I even do these?

Mathematics
1 answer:
Nady [450]2 years ago
6 0

Answer:

0?8

Step-by-step explanation:

Cos(A) = Adjacent/hypotenuse

Cos(A) = 36/45

=0.8

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In a laboratory, they were testing a certain bacteria. They stated with 50 bacteria and they noticed it triples every 30 minutes
MaRussiya [10]
I think it’s just 4(50•3)=600
4 0
2 years ago
Simplify 3(2x^4*y^5)^2/(6x^3*y^3)^6
Thepotemich [5.8K]

\dfrac{3(2x^4y^5)^2}{(6x^3y^3)^6}\qquad\text{use}\ (ab)^n=a^nb^n\ \text{and}\ (a^n)^m=a^{nm}\\\\=\dfrac{3(2^2)(x^4)^2(y^5)^2}{6^6(x^3)^6(y^3)^6}=\dfrac{3(4)x^8y^{10}}{46656x^{18}y^{18}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\dfrac{12x^8y^{10}}{46656x^8x^{10}y^{10}y^8}=\dfrac{1}{3888x^{10}y^8}

7 0
3 years ago
Express p in terms of q if
Irina-Kira [14]

p=x^2+\dfrac{1}{x^2}\\\\p=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\\\p=\dfrac{x^4+1}{x^2}\qquad(*)

q=x+\dfrac{1}{x}\\\\q=\dfrac{x^2}{x}+\dfrac{1}{x}\\\\q=\dfrac{x^2+1}{x}\qquad\text{square both sides}\\\\q^2=\left(\dfrac{x^2+1}{x}\right)^2\\\\q^2=\dfrac{(x^2+1)^2}{x^2}\qquad\text{use}\ \ (a+b)^2=a^2+2ab+b^2\\\\q^2=\dfrac{(x^2)^2+2(x^2)(1)+1^2}{x^2}\\\\q^2=\dfrac{x^4+2x^2+1}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+\dfrac{2x^2}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+2\qquad\text{subtract 2 from both sides}\\\\q^2-2=\dfrac{x^4+1}{x^2}\\\\\text{From (*) we have}\\\\\boxed{p=q^2-2}

8 0
2 years ago
We can use the formula, density = mass over volume, to find the density of a substance. A square metal plate has a density of 10
mash [69]

Answer:

2.15cm^3

Step-by-step explanation:

V = Mass / Density

21.93g / 10.2g/cm^3

2.15cm^3

3 0
2 years ago
HELPP ASaPP I’ll mark you as brainlister
nika2105 [10]

Answer:

BC ≈ 8.9 units

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos36° = \frac{adjacent}{hypotenuse} = \frac{BC}{AB} = \frac{BC}{11} ( multiply both sides by 11 )

11 × cos36° = BC , then

BC ≈ 8.9 ( to the nearest tenth )

5 0
2 years ago
Read 2 more answers
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