Write a slope-intercept equation for a line that passes through ( - (4,5) and (4,

Describe the graph of the quadratic from the original y = x^2 .

Citrus2011 [14]

Answer:

a. Narrower

b. Shifts left

c. Opens up

d. Shifts up

Step-by-step explanation:

The original quadratic equation is y = x²

The given quadratic equation is y = 5·(x + 4)² + 7

The given quadratic equation is of the form, f(x) = a·(x - h)² + k

a. A quadratic equation is narrower than the standard form when the coefficient is larger than the coefficient in the original equation

The quadratic coefficient is 5 > 1 in the original, therefore, the quadratic equation is narrower

b. Given that the given quadratic equation has positive 'a', and 'b', and h = -4, therefore, the axis of symmetry shifts left

c. The quadratic coefficient is positive, (a = 5), therefore, the quadratic equation opens down

d. The value of 'k' gives the vertical shift, therefore, the given quadratic equation with k = 7, shifts up.

4 0

2 years ago

8. Which of the numbers in each pair is farther to the left on the number line?

galben [10]

A. 305
b. 900
c. 46
d. 157,019

6 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

The top question plezzzzzzz zzzzzzzzz zzzzzzzzz zzzzzzzzz

zaharov [31]

Use a ruler for this problem. Take a picture of the measurement & then I could help you.

8 0

3 years ago

What is the inverse of the function f(x) = 2x – 10?

noname [10]

Answer:

The inverse f-1(x) = ( x + 10)/2.

Step-by-step explanation:

f(x) = 2x - 10

Find x in terms of f(x):

2x = f(x) + 10

x = (f(x) + 10))/ 2

Replace x by the inverse f-1(x) and f(x) by x:

f-1(x) = (x + 10)/2.

7 0

3 years ago

Write a slope-intercept equation for a line that passes through ( - (4,5) and (4, - 11).