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Rudik [331]
3 years ago
10

Find a gradient of a line that is parallel and perpendicular to this line with this gradient of -2

Mathematics
1 answer:
yulyashka [42]3 years ago
7 0

a gradient of a line that is parallel and perpendicular to this line with this gradient of -2

Gradient is the slope

So slope of the line =-2

Slope of parallel line is equal to the slope of the line

So slope of parallel line = -2

Slope of perpendicular line is equal to negative reciprocal of slope of the line

We know slope of line = -2

Negative reciprocal = \frac{1}{2}

So , Slope of perpendicular line= \frac{1}{2}

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nikitadnepr [17]
Scale of the model = 3.5 / 10.5 = 1/3
so the length of the model = 1/3  * 12 = 5 feet

Area of the scale model = 4 * 3.5 = 14 sq ft
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3 years ago
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A random sample of pickles in a jar has the following weights, in grams: 68, 54, 59, 70, 69, 65, 73, 59, 74, and 54. What is the
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All the weight of the pickles or the mean is 
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Which point is the solution to the following system of equations?
Naya [18.7K]

The point (3, 2) is the solution to given system of equations

<em><u>Solution:</u></em>

Given that system of equations are:

x^2 + y^2 = 13    ------ eqn 1

2x - y = 4    ------- eqn 2

From eqn 2,

y = 2x - 4

Substitute y = 2x - 4 in eqn 1

x^2 + (2x - 4)^2 = 13\\\\x^2 + 4x^2 + 16 - 16x = 13\\\\5x^2 -16x + 3 = 0

Let us solve the above equation by quadratic formula,

\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Using the Quadratic Formula for 5x^2 -16x + 3 = 0 where  a = 5, b = -16, and c = 3

\begin{aligned}&x=\frac{-(-16) \pm \sqrt{(-16)^{2}-4(5)(3)}}{2 \times 5}\\\\&x=\frac{16 \pm \sqrt{256-60}}{10}\\\\&x=\frac{16 \pm \sqrt{196}}{10}\end{aligned}

The discriminant b^2 - 4ac>0 so, there are two real roots.

\begin{aligned}&x=\frac{16 \pm \sqrt{196}}{10}=\frac{16 \pm 14}{10}\\\\&x=\frac{16+14}{10} \text { or } \frac{16-14}{10}\\\\&x=\frac{30}{10} \text { or } x=\frac{2}{10}\\\\&x=3 \text { or } x=0.2\end{aligned}

Substitute for x = 0.2 and x = 3 in 2x - y = 4

<em><u>when x = 3</u></em>

2(3) - y = 4

6 - y = 4

y = 2

<em><u>when x = 0.2</u></em>

2(0.2) - y = 4

0.4 - y = 4

y = 0.4 - 4

y = -3.6

Thus Option D is correct The point is (3, 2)

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Step-by-step explanation:

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