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2 years ago

The width of a rectangle is y feet long. The rectangle is 4 feet longer than it is wide. What is the area of the rectangle?

Mathematics

2 answers:

Ainat [17]2 years ago

8 0

Y(y+4) sq ft is the answer...

Dmitriy789 [7]2 years ago

5 0

Answer:

The answer is a. y(y + 4) sq ft

Step-by-step explanation:

the formula for the area of a rectangle is l*w

the y on the outside represents the width and the stuff inside the ( ) represent the length

so if you multiply them, you get the area

please mark as brainliest

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Initially, there were only 86 weeds in the garden. The weeds grew at a rate of 29% each week. The following function represents

Dmitry [639]

Answer:

Option A. $f(x)=86[1.04]^{x}$ ; grows approximately at a rate of 0.4% daily

Step-by-step explanation:

we have

$f(x)=86(1.29)^{x}$

where

f(x) the number of weeds in the garden

x ----> the number of weeks

Calculate how quickly the weeds grow each day

Remember that a week is equal to seven days

$f(x)=86(1.29)^{\frac{x}{7}}$

Using the law of exponents

b^(x/a) = b^(x*(1/a)) = (b^(1/a))^x

$f(x)=86[(1.29)^{\frac{1}{7}}]^{x}$

$f(x)=86[1.04]^{x}$

therefore

The rate is approximately

1.04=1+r

r=1.04-1=0.04=4% daily

8 0

2 years ago

What are the solutions to the equation

frosja888 [35]

Answer:

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

Step-by-step explanation:

You have the quadratic function $2x^2-x+1=0$ to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions $ax^2+bx+c=0$ with $a\neq 0$ the Bhaskara's Formula is:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}$

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}$

It usually has two solutions.

Then we have $2x^2-x+1=0$ where a=2, b=-1 and c=1. Applying the formula:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}$

Observation: $\sqrt{-1}=i$

$x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$

And,

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i$

Then the correct answer is option C.

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

3 0

2 years ago

Building engineer analyzes a concrete column with a circular cross section. The circumference of the column is 18 \pi18π18, pi m

Ksivusya [100]

the area A of the cross section of the column is $(81\pi )m^2$ .

<u>Step-by-step explanation:</u>

Here we have , building engineer analyzes a concrete column with a circular cross section. The circumference of the column is 18π, pi meters. We need to find What is the area A of the cross section of the column .Let's find out:

We know that , Circumference of circle = $2\pi r$