What is this equation rewritten in logarithmic form?

Mr peterson will buy 80 feet of fencing. The pieces of fence come in whole numbers of feet. His enclosure will be a rectangle an

DIA [1.3K]

The perimeter is:
P = 2x + 2y = 80
The area is:
A = x * y
The area as a function of a variable is:
A (x) = x * (40-x)
Rewriting we have:
A (x) = 40x-x ^ 2
We derive the function:
A '(x) = 40-2x
We equal zero and clear x:
40-2x = 0
2x = 40
x = 40/2
x = 20 feet
The other dimension is:
y = 40 - x
y = 40 - 20
y = 20 feet
The area will be:
A = x * y
A = 20 * 20
A = 400 feet ^ 2
Answer:
The dimensions are:
x = 20 feet
y = 20 feet
The area is:
A = 400 feet ^ 2

5 0

3 years ago

I need help rn PLEASEEE heheheh

snow_lady [41]

Diameter = 51 cm
Radius (r) = 51 cm ÷ 2 = 25.5 cm
π = 3.14
Circumference
= 2πr
= 2 × 3.14 × 25.5 cm
= 160.14 cm
Area of a circle
= πr^2
= 3.14 × (25.5)^2 cm^2
= 3.14 × 25.5 × 25.5 cm^2
= 2041.785 cm^2

Hope you could understand.

If you have any query, feel free to ask.

4 0

2 years ago

Which describes how to calculate the range of this data set?

ELEN [110]

Answer:

B) Subtract 12-4 to calculate the range of this data set

5 0

3 years ago

Find an e equation of the vertical line that passes through (0,-2)

viktelen [127]

The equation of the vertical line that satisfies the condition is x=0.

4 0

3 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .

Subtract $-12x^2 -36x$ from both sides of the equation:

$38x + 7 = c(x + 3)$ .

By the same logic, $c = 38$ .

Hence the quotient would be $(4x^2 - 12x + 38)$ .

6 0

3 years ago