All categories

3 years ago

What is the equation, in point-slope form, for a line that goes through(8,

Mathematics

1 answer:

LUCKY_DIMON [66]3 years ago

8 0

Answer:

$\boxed{y-4=-\frac{5}{6}(x-8)}$

Step-by-step explanation:

Point slope form is: $y-y_1=m(x-x_1)$

$(x_1,y_1) - Coordinate\\ m- Slope$

We are given the slope of $-\frac{5}{6}$ and the point of (8,4).

Replace 'm' with -5/6, 'x1' with 8, and 'y1' with 4.

$y-y_1=m(x-x_1)\rightarrow \boxed{y-4=-\frac{5}{6}(x-8)}$

I do not see any options that match this, but the boxed equation should the correct answer.

You might be interested in

Find the area of the quadrilateral ABCD.

REY [17]

Answer
—————————————-

3 0

2 years ago

What are the exact solutions of x2 − 5x − 7 = 0, where x equals negative b plus or minus the square root of b squared minus 4 ti

Marat540 [252]

Answer:

$x = \frac{5 + \sqrt{53}}{2}$ or $x = \frac{5 - \sqrt{53}}{2}$

Step-by-step explanation:

Given

$x^2 - 5x - 7 = 0$

Required

Solve for x using:

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

First, we need to identify a, b and c

The general form of a quadratic equation is:

$ax^2 + bx + c = 0$

So, by comparison with $x^2 - 5x - 7 = 0$

$a = 1$ $b = -5$ $c = -7$

Substitute these values of a, b and c in

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-(-5) \± \sqrt{(-5)^2 - 4 * 1 * -7}}{2 * 1}$

$x = \frac{5 \± \sqrt{25 +28}}{2}$

$x = \frac{5 \± \sqrt{53}}{2}$

Split the expression to two

$x = \frac{5 + \sqrt{53}}{2}$ or $x = \frac{5 - \sqrt{53}}{2}$

To solve further in decimal form, we have

$x = \frac{5 + 7.28}{2}$ or $x = \frac{5 - 7.28}{2}$

$x = \frac{12.28}{2}$ or $x = \frac{-2.28}{2}$

$x = 6.14$ or $x = -1.14$

4 0

3 years ago

Write 42/100 in standard form as a decimal

iragen [17]

Answer:

0.42

Step-by-step explanation:

divide 100÷42 then the answer is 0.42

3 0

3 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago

Helppppppppppppppppppppppppppp

Wittaler [7]

It will be itself because 1 cannot be divided by the number itself

4 0

2 years ago