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3 years ago

Write the equation of the line shown in point slope form (1,2) (3,6)

Mathematics

1 answer:

Kipish [7]3 years ago

7 0

The point-slope form:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (1, 2) and (3, 6). Substitute:

$m=\dfrac{6-2}{3-1}=\dfrac{4}{2}=2\\\\\boxed{y-2=2(x-1)}$

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kap26 [50]

Answer:

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Step-by-step explanation:

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2 years ago

-4/5 : 1/3 I need help with this

Brums [2.3K]

Answer:

Step-by-step explanation:

-4/5÷1/3= (-4/5)x(3/1)= -12/5=-2 2/5

6 0

3 years ago

Given the function LaTeX: f\left(x\right)=-5x^2-x+20f ( x ) = − 5 x 2 − x + 20, find LaTeX: f\left(2\right)f ( 2 ). Group of ans

BartSMP [9]

The value of f(2) is -2 ⇒ 1st answer

Step-by-step explanation:

The form of the quadratic function is;

f(x) = ax² + bx + c, where

a is the coefficient of x²
b is the coefficient of x
c is the numerical term
x is the domain of the function and f(x) is the range of the function

∵ f(x) = -5x² - x + 20

- f(2) means value f(x) at x = 2

∵ x = 2

- Substitute x in the function by 2 to find f(2)

∵ f(2) = -5(2)² - (2) + 20

∴ f(2) = -5(4) - 2 + 20

∴ f(2) = -20 - 2 + 20

∴ f(2) = -2

The value of f(2) is -2

Learn more:

You can learn more about the function in brainly.com/question/12363217

#LearnwithBrainly

6 0

2 years ago

I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit

makkiz [27]

Answer:

Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

Step-by-step explanation:

Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

If the denominator of $c$ is also $2$ , then the range for its numerator (call it $p$ ) would be $3 < p < 4$ . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than $2$ .

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

$\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}$ .

$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

$\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}$ .

$\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}$ .

$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .