I need help with graph the question is .

Change this from vertex form to standard form

wolverine [178]

Answer:

y=4x^2-24x+41

hope this helps!

mark brainliest?

3 0

2 years ago

Anyone know how to solve this?

yuradex [85]

Answer:

Step-by-step explanation:

<h3> $( - \frac{4}{5})^{2}$ </h3><h3> ${( - 1)}^{2} \frac{ {4}^{2} }{ {5}^{2} }$ </h3><h3> $1 \frac{ {4}^{2} }{ {5}^{2} }$ </h3><h3> $\frac{ {4}^{2} }{ {5}^{2} }$ </h3><h3> $\frac{16}{25}$ </h3><h3>Hope it is helpful....</h3>

5 0

3 years ago

Bradley and Kelly are out flying kites at a park one afternoon. A model of Bradley and Kelly's kites are shown below on the coor

Alex787 [66]

The correct answer is:

C. They are similar because the corresponding sides of kites KELY and BRAD all have the relationship 2:1.

Using the distance formula,

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

the lengths of the sides of BRAD are:

$\text{BR}=\sqrt{(7-6)^2+(3-4)^2}=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\\\\\text{RA}=\sqrt{(6-3)^2+(4-3)^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{AD}=\sqrt{(3-6)^2+(3-2)^2}=\sqrt{(-3)^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{DB}=\sqrt{(6-7)^2+(2-3)^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}$

The lengths of the sides of KELY are:

$\text{KE}=\sqrt{(2-0)^2+(11-9)^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\\\\\text{EL}=\sqrt{(0-2)^2+(9-3)^2}=\sqrt{(-2)^2+6^2}=\sqrt{40}=2\sqrt{10}\\\\\text{LY}=\sqrt{(2-4)^2+(3-9)^2}=\sqrt{(-2)^2+(-6)^2}=\sqrt{40}=2\sqrt{10}\\\\\text{YK}=\sqrt{(4-2)^2+(9-11)^2}=\sqrt{2^2+(-2)^2}=\sqrt{8}=2\sqrt{2}$