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3 years ago

What function would represent a translation of this graph of y=x^2 by 8 units to the right?

Mathematics

1 answer:

Anit [1.1K]3 years ago

5 0

Answer:

$y = (x-8)^2$

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Pls help to get brainlist

Lubov Fominskaja [6]

Answer: Please give brainliest!!!

$\frac{3}{2} = \frac{15}{10}$

$\frac{15}{10} - \frac{12}{10} = \frac{3}{10}$

4 0

2 years ago

Read 2 more answers

Evaluate −8x+5−2x−4+5x when x=2

ollegr [7]

−8x+5−2x−4+5x

-8(2)+5-2(2)-4+5(2)

(-16)+5-4-4+10

-11+5-4-4+10

-6+5-4-4+10

-1+5-4-4+10

4-4-4+10

0-4+10

-4+10

Answer: 6

4 0

3 years ago

Neil has 21 pizzas. He serves 3 1/2 pizzas to his family for dinner. What fraction of the pizzas is left?

Ket [755]

Answer:

17 1/2

Step-by-step explanation:

21 - 3 1/2 = 17 1/2

5 0

2 years ago

Read 2 more answers

A triangle is graphed in the coordinate plane. The vertices of the triangle have coordinates (–3, 1), (1, 1), and (1, –2). What

vladimir2022 [97]

Answer:

The perimeter of the triangle is $12\ units$

Step-by-step explanation:

Let

$A(-3,1),B(1,1),C(1,-2)$

we know that

The perimeter of triangle is equal to

$P=AB+BC+AC$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

$A(-3,1),B(1,1)$

substitute in the formula

$AB=\sqrt{(1-1)^{2}+(1+3)^{2}}$

$AB=\sqrt{(0)^{2}+(4)^{2}}$

$AB=4\ units$

step 2

Find the distance BC

$B(1,1),C(1,-2)$

substitute in the formula

$BC=\sqrt{(-2-1)^{2}+(1-1)^{2}}$

$BC=\sqrt{(-3)^{2}+(0)^{2}}$

$BC=3\ units$

step 3

Find the distance AC

$A(-3,1),C(1,-2)$

substitute in the formula

$AC=\sqrt{(-2-1)^{2}+(1+3)^{2}}$

$AC=\sqrt{(-3)^{2}+(4)^{2}}$

$AC=5\ units$

step 4

Find the perimeter

$P=AB+BC+AC$

substitute the values

$P=4+3+5=12\ units$

6 0

3 years ago

Sistema de ecuaciones.5x+2y=-152x-2y=-6

Zolol [24]

Tnemos el sisema de ecuaciones:

$\begin{gathered} 5x+2y=-15 \\ 2x-2y=-6 \end{gathered}$

Podemos resolverlo por eliminación sumando ambas ecuaciones y eliminando y. Asi podemos resolver para x:

$\begin{gathered} (5x+2y)+(2x-2y)=(-15)+(-6) \\ 7x+0y=-21 \\ x=-\frac{21}{7} \\ x=-3 \end{gathered}$

Ahora podemos resolver para y con cualquiera de las dos ecuaciones:

$\begin{gathered} 2x-2y=-6 \\ 2\cdot(-3)-2y=-6 \\ -6-2y=-6 \\ -2y=-6+6 \\ -2y=0 \\ y=0 \end{gathered}$

Respuesta: x=-3, y=0

7 0

10 months ago