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3 years ago

Manuel is building a circular pool with an area of 100 square feet he will build rectangular deck around the pool as shown

Mathematics

2 answers:

liberstina [14]3 years ago

8 0

Answer:

length of a rectangle deck = l = 26 square

width of a rectangle deck = w = 22 square

Step-by-step explanation:

Let,

length of a rectangle deck = l

width of a rectangle deck = w

Given Data:

Area of the deck = 572 square feet

According to given condition;

w = l - 4

To Find Out:

length of a rectangle deck = i = ?

width of a rectangle deck = w = ?

Formula:

Area of a rectangle = A = l × w

Solution:

Area of a rectangle = A = l × w

572 = l × w

572 = l × ( l - 4)

$572 = l^{2} - 4l$

$l^{2} - 4l - 572 = 0$

$l^{2} - 26l + 22l - 572 = 0$

$(l^{2} - 26l) + (22l - 572) = 0$

$l(l - 26) + 22(l - 26) = 0$

$(l - 26) (l + 22) = 0$

$l - 26=0 \\$ or $l+22=0$

$l =26 \\$ or $l=-22$

As length is always positive so, $l =26 \\$ square

w = 26 - 4

w = 22 square

kirill [66]3 years ago

7 0

Answer:

26 feet

Step-by-step explanation:

w: width

w+4 : length

w(w+4) = 572

w² + 4w - 572 = 0

w² + 26w - 22w - 572 = 0

w(w+26) - 22(w+26) = 0

(w-22)(w+26) = 0

w = 22, -26

width = 22 feet

length = w+4 = 22+4 = 26 feet

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4 years ago

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Blababa [14]

Answer:

If you mean only one rational solution, the answer is

$k_1 = 8, k_2 = -8$

If you mean at least 1 rational solution, the answer is

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Step-by-step explanation:

$4x^2 + kx + 4 = 0$

Let's calculate the discriminant.

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$\Delta = k^2 -4 \cdot 4 \cdot 4$

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Now, remember that:

$\text{If } \Delta > 0 : \text{2 Real solutions}$

$\text{If } \Delta = 0 : \text{1 Real solution}$

$\text{If } \Delta < 0 : \text{No Real solution}$

Therefore, I will just consider the first two cases.

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and

$k^2 -64 = 0$

$\boxed{\text{For } k^2 - 64 > 0}$

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The Speedmaster IV automobile gets an average of 22.0 miles per gallon in the city. The standard deviation is 3 miles per gallon

aniked [119]

Answer:

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Think of (or draw) the standard normal curve. Mark the mean (22.0). Then one standard deviation above the mean would be 22.0 + 3.0, or 25.0. Two would be 22.0 + 2(3.0), or 28.0. Finallyl, draw a vertical line at 26.0.

Our task is to determine the area under the curve to the left of 26.0.

Using a basic calculator with built-in statistical functions, we find this area as follows:

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