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4 years ago

Determinet the x-intercept for 3x+2y=14

Mathematics

1 answer:

DerKrebs [107]4 years ago

3 0

The x-intercept is for y = 0.

Put y = 0 to the equation of a line 3x + 2y = 14

$3x+2(0)=14\\\\3x=14\qqud\text{divide both sides by 3}\\\\\boxed{x=\dfrac{14}{3}\to\left(\dfrac{14}{3};\ 0\right)}$

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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

1 year ago

Please help. I am doing a test and I do not understand what I am being asked.

Oksanka [162]

Answer:

one

Step-by-step explanation:

i have no idea what kind of test ur taking my guy but have fun

5 0

3 years ago

Find x so that l || m. state the converse used.

kenny6666 [7]

Answer:

$x = 12$

Converse: Alternate Interior Angles Converse

Step-by-step explanation:

By the Alternate Interior Angles Converse, if $(14x - 23) = (9x + 37)$ , then l || m.

Use the equation to solve for x as follows:

$14x - 23 = 9x + 37$

Subtract 9x from each side

$14x - 23 - 9x = 9x + 37 - 9x$

$5x - 23 = 37$

Add 23 to both sides

$5x - 23 + 23 = 37 + 23$

$5x = 60$

Divide both sides by 5

$\frac{5x}{5} = \frac{60}{5}$

$x = 12$

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3 years ago

1/2x = -40<br> What is x?

Sonbull [250]

Answer:

x = -80

Explanation

-40 divided by 1/2 is -80

5 0

3 years ago

Read 2 more answers

Find the area under the standard normal probability distribution between the following pairs of z-scores. a. z=0 and z=3.00 e.

prohojiy [21]

Answer:

a. P(0 < z < 3.00) = 0.4987

b. P(0 < z < 1.00) = 0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of z-scores.

a. z=0 and z=3.00