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3 years ago

Which equation of the line written in standard form has has a slope of 6 and contains the points (-1,-3)

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

5 0

keeping in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

$\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-3})~\hspace{10em} slope = m\implies 6 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-3)=6[x-(-1)]\implies y+3=6(x+1) \\\\\\ y+3=6x+6\implies y=6x+3\implies -6x+y=3\implies 6x-y=-3$

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Final answer is 10

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Please help find domain and range

Oxana [17]

<h3>Answer: Choice A</h3>

Domain: x > 4
Range: y > 0

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Explanation:

We want to avoid having a negative number under the square root. Solving $x-4 \ge 0$ leads to $x \ge 4$

So it appears the domain could involve x = 4 itself; however, if we tried that x value, then we'd get a division by zero error.

So in reality, the domain is x > 4.

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The range of y = sqrt(x) is the set of positive real numbers. So y > 0 is the range for this equation. Shifting left and right does not affect the range, so the range of y = sqrt(x-4) is also y > 0.

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Therefore, the range of the given equation is y > 0

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The graph is shown below. We have a vertical asymptote at x = 4 and a horizontal asymptote at y = 0. The green curve is fenced in the upper right corner (northeast corner).

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It's not correct

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3 years ago

find the probability of being delt 5 clubs and 3 cards with one of each remaining suit in 8 card poker

kumpel [21]

Answer: 0.003757(approx).

Step-by-step explanation:

Total number of combinations of selecting r things out of n things is given by:-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Total cards in a deck = 52

Total number of ways of choosing 8 cards out of 52 = $^{52}C_8$

Total number of ways to choose 5 clubs and 3 cards with one of each remaining suit = $^{13}C_5\times^{13}C_1\times^{13}C_1\times^{13}C_1$ [since 1 suit has 13 cards]

The required probability = $=\dfrac{^{13}C_5\times^{13}C_1\times^{13}C_1\times^{13}C_1}{^{52}C_8}$

$=\dfrac{\dfrac{13!}{5!8!}\times13\times13\times13}{\dfrac{52!}{8!44!}}\\\\=\dfrac{24167}{6431950}\\\\\approx0.003757$

Hence, the required probability is 0.003757 (approx).

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3 years ago