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2 years ago

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Mathematics

1 answer:

Lemur [1.5K]2 years ago

8 0

Answer:

B. 74 / 100

Explanation:

I got it right on the test :)

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Solve for x. Round to the nearest tenth if necessary.

Natasha2012 [34]

Answer:

x = 16

Step-by-step explanation:

9/12 = 12/x

9x = 144

x = 16

7 0

3 years ago

Read 2 more answers

What is the equation of this graphed line?

Veseljchak [2.6K]

Answer:

Slope Intercept form of the equation is $y = \frac{1}{3} x - 9$

Step-by-step explanation:

Here, the two point line are given as is A(-6,-3) and B(6,-7)

The slope of the line AB = $m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{-7-(-3)}{6-(-6)} = \frac{-7 + 3}{6 + 6} = \frac{4}{12} \\\implies m = \frac{1}{3}$

⇒ the slope of AB is m = (4/3)

By SLOPE INTERCEPT FORMULA:

The equation of a line with slope m and a point (x0, y0) is given as

(y-y0)= m (x-x0)

⇒ The equation of line with point (6,-7) is:

$y + 7 = \frac{1}{3} (x-6) \implies 3y + 21 - x + 6 - 0\\or, -x + 3y + 27 = 0$

Now, the given equation is -x + 3y = -27

Convert it in the SLOPE INTERCEPT FORM y = mx + c

We get, 3y = x - 27

or, $y = \frac{1}{3} x - \frac{27}{3} \\\implies y = \frac{1}{3} x - 9$

Hence, the Slope Intercept form of the equation is $y = \frac{1}{3} x - 9$

3 0

2 years ago

Find a quadratic polynomial with integer coefficients which has x = 2/9, ± SQRT23/9 as its real zeros.

melomori [17]

Answer:

The quadratic polynomial with integer coefficients is $y = 81\cdot x^{2}-36\cdot x -19$ .

Step-by-step explanation:

Statement is incorrectly written. Correct form is described below:

<em>Find a quadratic polynomial with integer coefficients which has the following real zeros: </em> $x = \frac{2}{9}\pm \frac{\sqrt{23}}{9}$ <em>. </em>

Let be $r_{1} = \frac{2}{9}+\frac{\sqrt{23}}{9}$ and $r_{2} = \frac{2}{9}-\frac{\sqrt{23}}{9}$ roots of the quadratic function. By Algebra we know that: