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1 year ago

Find an equation of each line given one point and the slope : (4, 8); slope 8

Mathematics

1 answer:

Fudgin [204]1 year ago

3 0

Step-by-step explanation:

in such a case the easiest approach might be using the point-slope form of a line equation :

y - y1 = m(x - x1)

with m being the slope, and (x1, y1) being a point on the line.

so, in our case that would be

y - 8 = 8(x - 4)

simplified this would be

y - 8 = 8x - 32

y = 8x - 24

FYI - this would be then the slope-intercept form (-24 would be the intercept point on the y-axis).

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What is HCF and LCM ? # confused

GaryK [48]

HCF= highest common factor
LCM= least common multiple

7 0

2 years ago

Read 2 more answers

Which graph shows a line with an x-intercept of -5?

Finger [1]

Answer:

The answer is A.

Step-by-step explanation:

The only option is A. since an intercept of (-5,0)

Option B has y-int: (0,-5)
Option C has y-int: (0,5)

Option D has x-int(5,0)

8 0

1 year ago

A sample of 10 adult elephants had an average weight of 12,556 pounds. The standard deviation

kirill [66]

Answer:

The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 10 - 1 = 9

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 3.25

The margin of error is:

M = T*s = 3.25*25 = 81

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 12,556 - 81 = 12,475 pounds

The upper end of the interval is the sample mean added to M. So it is 12,556 + 81 = 12,637 pounds.

The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.

6 0

2 years ago

Complete the square and write in standard form. Show all work.What would be the conic section:CircleEllipseHyperbolaParabola

mote1985 [20]

ANSWER

This is an ellipse. The equation is:

$\frac{(x-1)^2}{3^2}+\frac{(y+4)^2}{4^2}=1$

EXPLANATION

We have to complete the square for each variable. To do so, we have to take the first two terms and compare them with the perfect binomial squared formula,

$(a+b)^2=a^2+2ab+b^2$

For x we have to take 16x² and -32x. Since the coefficient of x is not 1, first, we have to factor out the coefficient 16,

$16x^2-32x=16(x^2-2x)$

Now, the first term of the expanded binomial would be x and the second term -2x. Thus, the binomial is,

$(x-1)^2=x^2-2x+1$

To maintain the equation, we have to subtract 1,

$16(x^2-2x+1-1)=16((x-1)^2-1)=16(x-1)^2-16$

Now, we replace (16x² - 32x) from the given equation by this equivalent expression,

$16(x-1)^2-16+9y^2+72y+16=0$

The next step is to do the same for y. We have the terms 9y² + 72y. Again, since the coefficient of y² is not 1, we factor out the coefficient 9,

$9y^2+72y=9(y^2+8y)$

Following the same reasoning as before, we have that the perfect binomial squared is,

$(y+4)^2=y^2+8y+16$

Remember to subtract the independent term to maintain the equation,

$9(y^2+8y)=9(y^2+8y+16-16)=9((y+4)^2-16)=9(y+4)^2-144$

And now, as we did for x, replace the two terms (9y² + 72y) with this equivalent expression in the equation,

$16(x-1)^2-16+9(y+4)^2-144+16=0$

Add like terms,

$\begin{gathered} 16(x-1)^2+9(y+4)^2+(-16-144+16)=0 \\ 16(x-1)^2+9(y+4)^2-144=0 \end{gathered}$

Add 144 to both sides,

$\begin{gathered} 16(x-1)^2+9(y+4)^2-144+144=0+144 \\ 16(x-1)^2+9(y+4)^2=144 \end{gathered}$

As we can see, this is the equation of an ellipse. Its standard form is,

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

So the next step is to divide both sides by 144 and also write the coefficients as fractions in the denominator,

$\begin{gathered} \frac{16(x-1)^2}{144}+\frac{9(y+4)^2}{144}=\frac{144}{144} \\ \\ \frac{(x-1)^2}{\frac{144}{16}}+\frac{(y+4)^2}{\frac{144}{9}}=1 \end{gathered}$

Finally, we have to write the denominators as perfect squares, so we identify the values of a and b. 144 is 12², 16 is 4² and 9 is 3²,

$\frac{(x-1)^2}{(\frac{12}{4})^2}+\frac{(y+4)^2}{(\frac{12}{3})^2}=1$

Note that we can simplify a and b,

$\frac{12}{4}=3\text{ and }\frac{12}{3}=4$

Hence, the equation of the ellipse is,

$\frac{(x-1)^2}{3^2}+\frac{(y+4)^2}{4^2}=1$

3 0

11 months ago

If the factors of function f are (x - 6) and (x - 1), what are the zeros of function f?

lidiya [134]

X=6 and x=1

You can find your zeros by determining what you have to plug into the function in order for it to equal zero

If we plug in 6, for example we’d get (6-6)(x-1)

Simplified this is 0(x-1)

Anything times 0 is 0, so this is one of our zeros.

Same goes for x-1, we just need to plug in 1 for it to equal 0

Therefore there are zeros at x=1 and x=6 :))

7 0

2 years ago