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3 years ago

Can someone please help me with these 3 questions I’ll mark brainliest

Mathematics

1 answer:

pshichka [43]3 years ago

7 0

Answer:

2/ triple need video for the third answer

Step-by-step explanation:

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Which statement is true about the local minimum of the

oksano4ka [1.4K]

Answer: D

Over the interval [4,7], the local minimum is -7

5 0

3 years ago

Write the number in 2 equivalent forms as a fraction, decimal, or percent. 11 16

pashok25 [27]

11.
Fraction - 11/100
Decimal - 0.11
Percent - 11%
16.
Fraction - 16/100
Decimal - 0.16
Percent - 16%

3 0

3 years ago

A rectangular parking lot has an area of 15,000 feet squared, the length is 20 feet more than the width. Find the dimensions

faust18 [17]

Dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet

<h3>Solution:</h3>

Given that

Area of rectangular parking lot = 15000 square feet

Length is 20 feet more than the width.

Need to find the dimensions of rectangular parking lot.

Let assume width of the rectangular parking lot in feet be represented by variable "x"

As Length is 20 feet more than the width,

so length of rectangular parking plot = 20 + width of the rectangular parking plot

=> length of rectangular parking plot = 20 + x = x + 20

The area of rectangle is given as:

$\text {Area of rectangle }=length \times width$

Area of rectangular parking lot = length of rectangular parking plot $\times$ width of the rectangular parking

$\begin{array}{l}{=(x+20) \times (x)} \\\\ {\Rightarrow \text { Area of rectangular parking lot }=x^{2}+20 x}\end{array}$

But it is given that Area of rectangular parking lot = 15000 square feet

$\begin{array}{l}{=>x^{2}+20 x=15000} \\\\ {=>x^{2}+20 x-15000=0}\end{array}$

Solving the above quadratic equation using quadratic formula

General form of quadratic equation is

${ax^{2}+\mathrm{b} x+\mathrm{c}=0$

And quadratic formula for getting roots of quadratic equation is

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

In our case b = 20, a = 1 and c = -15000

Calculating roots of the equation we get

$\begin{array}{l}{x=\frac{-(20) \pm \sqrt{(20)^{2}-4(1)(-15000)}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{400+60000}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{60400}}{2}} \\\\ {x=\frac{-(20) \pm 245.764}{2 \times 1}}\end{array}$

$\begin{array}{l}{=>x=\frac{-(20)+245.764}{2 \times 1} \text { or } x=\frac{-(20)-245.764}{2 \times 1}} \\\\ {=>x=\frac{225.764}{2} \text { or } x=\frac{-265.764}{2}} \\\\ {=>x=112.882 \text { or } x=-132.882}\end{array}$

As variable x represents width of the rectangular parking lot, it cannot be negative.

=> Width of the rectangular parking lot "x" = 112.882 feet

=> Length of the rectangular parking lot = x + 20 = 112.882 + 20 = 132.882

Hence can conclude that dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet.

3 0

3 years ago

2x-3y=42 solve for y

PtichkaEL [24]

Answer:

y=14+(2/3x)

Step-by-step explanation:

Your main goal is to get Y by itself on one side.

To start subtract 2x from both sides. This will give you -3y=42-2x.

Then you divide both sides by -3 to get Y by itself.

That will leave you with ...

y=14+(2/3x)

(remember the negatives cancel each other out)

5 0

3 years ago

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Algebra one quiz 12.2.3

Montano1993 [528]

Answer:

where is link

Step-by-step explanation:

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3 years ago

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