Help me please I need the help

Derrick and Sam need 10 pounds of colored gravel for the bottom of their new fish tank. Derrick has 2 pounds of purple gravel an

Sunny_sXe [5.5K]

Answer:

4 yellow gravel . 2+4=6

10-6= 4

8 0

3 years ago

Two more than four times a number is three

klio [65]

Answer:

1/4 is the number

a.k.a. 0.25

Step-by-step explanation:

Replace the words "a number" with x

4x + 2 = 3

Then, simplify

4x = 1

x = 1/4

5 0

3 years ago

Find the value of the following expression:

Ket [755]

(2^8 . 5^-5 . 19^0)^-2

using PEMDAS

this = (2^8 * 1/5^5 * 1)^-2

= (5^5 / 2^8)^2

= 5^10 / 2^16

7 0

3 years ago

Read 2 more answers

3/5+1/3= ...............................................................................................................

Mrac [35]

Answer:

Step-by-step explanation:

you add two fractions:

3/5+1/3=

find the common denominator = it is 15 (3*5)

9/15+5/15=14/15

7 0

3 years ago

What would y=x^2 +x+ 2 be in vertex form

balu736 [363]

Answer:

$y = (x + \frac{1}{2} )^{2} + \frac{7}{4}$

Step-by-step explanation:

$y = {x}^{2} + x + 2$

We can covert the standard form into the vertex form by either using the formula, completing the square or with calculus.

$y = a(x - h)^{2} + k$

The following equation above is the vertex form of Quadratic Function.

Vertex — Formula

$h = - \frac{b}{2a} \\ k = \frac{4ac - {b}^{2} }{4a}$

We substitute the value of these terms from the standard form.

$y = a {x}^{2} + bx + c$

$h = - \frac{1}{2(1)} \\ h = - \frac{ 1}{2}$

Our h is - 1/2

$k = \frac{4(1)(2) - ( {1})^{2} }{4(1)} \\ k = \frac{8 - 1}{4} \\ k = \frac{7}{4}$

Our k is 7/4.

Vertex — Calculus

We can use differential or derivative to find the vertex as well.

$f(x) = a {x}^{n}$

Therefore our derivative of f(x) —

$f'(x) = n \times a {x}^{n - 1}$

From the standard form of the given equation.

$y = {x}^{2} + x + 2$

Differentiate the following equation. We can use the dy/dx symbol instead of f'(x) or y'

$f'(x) = (2 \times 1 {x}^{2 - 1} ) + (1 \times {x}^{1 - 1} ) + 0$

Any constants that are differentiated will automatically become 0.

$f'(x) = 2 {x}+ 1$

Then we substitute f'(x) = 0

$0 =2x + 1 \\ 2x + 1 = 0 \\ 2x = - 1 \\x = - \frac{1}{2}$

Because x = h. Therefore, h = - 1/2

Then substitute x = -1/2 in the function (not differentiated function)

$y = {x}^{2} + x + 2$

$y = ( - \frac{1}{2} )^{2} + ( - \frac{1}{2} ) + 2 \\ y = \frac{1}{4} - \frac{1}{2} + 2 \\ y = \frac{1}{4} - \frac{2}{4} + \frac{8}{4} \\ y = \frac{7}{4}$

Because y = k. Our k is 7/4.

From the vertex form, our vertex is at (h,k)

Therefore, substitute h = -1/2 and k = 7/4 in the equation.

$y = a {(x - h)}^{2} + k \\ y = (x - ( - \frac{1}{2} ))^{2} + \frac{7}{4} \\ y = (x + \frac{1}{2} )^{2} + \frac{7}{4}$

7 0

3 years ago