Can someone talk me through the steps?

Mathematics

1 answer:

dimulka [17.4K]3 years ago

3 0

Y = 6.9x + 5.1.....multiply by 10 to get rid of decimals
10y = 69x + 51 ...subtract 69x from both sides
-69x + 10y = 51

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What are the factors of 252 .

Naya [18.7K]

Answer:

252 is a composite number.

Prime factorization: 252 = 2 x 2 x 3 x 3 x 7, which can be written 252 = (2^2) x (3^2) x 7

The exponents in the prime factorization are 2, 2, and 1. Adding one to each and multiplying we get (2 + 1)(2 + 1)(1 + 1) = 3 x 3 x 2 = 18. Therefore 252 has 18 factors.

Factors of 252: 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252

Factor pairs: 252 = 1 x 252, 2 x 126, 3 x 84, 4 x 63, 6 x 42, 7 x 36, 9 x 28, 12 x 21, or 14 x 18

Taking the factor pair with the largest square number factor, we get √252 = (√7)(√36) = 6√7 ≈ 15.875

Step-by-step explanation:

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Lina20 [59]

Since 500 is the area, and 20 is the length, divide 500 by 20. Your answer is a decimal. (0.04) therefore, you have to make it into a whole number. So move the decimal to the right two times. So 4 is your height.

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Solve the equation by completing the square.Round to the nearest hundredth if necessary. X^2–x–7=0

lesya [120]

NOTE: $\left(x-\dfrac{1}{2}\right)^2=x^2-x+\dfrac{1}{4}$

$x^2-x-7=0 \\ \\ \underbrace{x^2-x+\frac{1}{4}}_{\left(x-\frac{1}{2}\right)^2}-7\dfrac{1}{4}=0 \\ \\ \\ \left(x-\frac{1}{2}\right)^2=7\frac{1}{4}=\frac{29}{4} \qquad /\sqrt{}\\ \\ \left|x-\frac{1}{2}\right|=\dfrac{\sqrt{29}}{2} \\ \\ x_1=\dfrac{1}{2}-\dfrac{\sqrt{29}}{2} \approx -2.19 \\ \\ x_2=\dfrac{1}{2}+\dfrac{\sqrt{29}}{2} \approx 3.19$

ANSWER: c)

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3 years ago

Let y=r(x) be continuous and differentiable. Assume r'(x)=0 r′(x)=0 at x=ax=a.

Basile [38]

Answer:

am in callege i need points really much

Step-by-step explanation:

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3 years ago

Solve the equation 2cosA = 3tanA

zavuch27 [327]

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\
-----------------------------\\\\

2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)}
\\\\\\
2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A)
\\\\\\
2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2$

$\bf \\\\\\
0=[2sin(A)-1][sin(A)+2]\implies 
\begin{cases}
0=2sin(A)-1\\
1=2sin(A)\\
\frac{1}{2}=sin(A)\\\\
sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\
\frac{\pi }{6},\frac{5\pi }{6}\\
----------\\
0=sin(A)+2\\
-2=sin(A)
\end{cases}$

now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1

now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine

so, only the first case are the valid angles for A

8 0

3 years ago