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3 years ago

Write simplified expressions for the area and perimeter of the rectangle.

Mathematics

1 answer:

mel-nik [20]3 years ago

6 0

Answer:

Area = 12x + 66

Perimeter = 2x + 35

Step-by-step explanation:

Area= 12(5.5 + x)

= 12x + 66

Perimeter = 2[(5.5 + x)+12]

= 2(17.5 + x)

= 2x + 35

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Rudik [331]

Answer:

a=1/2b, 2a=b

Step-by-step explanation:

Since angle a is 1/2 the size of angle b, a=1/2b and b=2a. In order to get a numeral answer, you need the number of one of the angles.

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2 years ago

2b^3+5=2(3)^3+5 when b=3

seraphim [82]

Answer:

2b^3+5=2(3)^3+5 when b=3

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3 years ago

The amount that results when $4,000 is compounded at 6% annually over seven years.

DerKrebs [107]

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Step-by-step explanation:

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3 years ago

-9f +19-1 + f) 12-=8f is this the answer?

jeka94

Hi!

-9f + 19 - 1 + f is actually -8f + 18.

To do this we add like terms, -9f and f; and 19 and -1

-9f + f is -8f

19 - 1 is 18

Therefore your answer is -8f + 18

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.

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

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3 years ago