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den301095 [7]
2 years ago
6

A smaller square of side length 17 feet is cut out of a square board. What is the approximate area (shaded region) of the remain

ing board in square feet?
Mathematics
1 answer:
gregori [183]2 years ago
7 0

Answer:

The area of the remaining board is (<em>x</em>² - 289) sq. ft.

Step-by-step explanation:

Let the sides of the bigger square board be, <em>x</em> feet.

It is provided that a smaller square of side length 17 feet is cut out of the bigger square board.

The area of a square is:

Area=(side)^{2}

Compute the area of the bigger square board as follows:

A_{b}=(side_{b})^{2}=x^{2}

Compute the area of the smaller square board as follows:

A_{s}=(side_{s})^{2}=(17)^{2}=289

Compute the area of the remaining board in square feet as follows:

\text{Remaining Area}=A_{b}-A_{s}

                          =[x^{2}-289]\ \text{square ft.}

Thus, the area of the remaining board is (<em>x</em>² - 289) sq. ft.

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The areas of the squares adjacent to two sides of a right triangle are 29.2529.2529, point, 25 units^2 2 squared and 131313 unit
bonufazy [111]

Answer:

6.5 units

Step-by-step explanation:

We are given that

Two adjacent sides of right triangle

Suppose a,b and c are sides of right triangle.

a^2=29.25squared units

b^2=13square units.

Length of third side,c=x units

We have to find the length of third side of the triangle.

x^2=a^2+b^2

By using Pythagoras theorem

(Hypotenuse)^2=(Perpendicular\;side)^2+(Base)^2

x^2=29.25+13

x=\sqrt{29.25+13}

x=6.5units

Hence, the length x of the third side of the triangle=6.5 units

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2 years ago
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Brainly to whoever is first and right:) - In quadrilateral ABCD, AB = BC = 1, m∠B=100°, m∠D=130°. Find BD
n200080 [17]

Answer: 1

B is the center of a circle of radius 1.  A and C are on the circle 100 degrees apart.  So arc AC is 100 degrees.  Arc AC the long way around is 360-100=260 degrees.   That means any point D on arc AC will subtend a 130 degree angle.  BD is a radius for all of those, and the radius is 1.

7 0
2 years ago
A weather balloon has a volume of 90.0 L when it is released from sea level 101 kPa. What is the atmospheric pressure on the bal
kirza4 [7]

Answer:

<em><u>The final atmospheric pressure is 5.19 · 10⁴ Pa</u></em>

Step-by-step explanation:

Assuming that the temperature of the air does not change, we can use Boyle's law, which states that for a gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume. In formula,

pV = const.

where p is the gas pressure and V is the volume

The equation can also be rewritten as

p₁ V₁ = p₂ V₂

where in our problem we have:

p₁ = 1.03 · 10₅ Pa is the initial pressure (the atmospheric pressure at sea level)

V₁ = 90.0L is the initial volume

p₂ is the final pressure

V₂ = 175.0L is the final volume

Solving the equation for p2, we find the final pressure:

p₂ = p₁ v₁ divided by V₂ = (1.01 · 10⁵)(90.0) divided by 175.0 = 5.19 · 10⁴ Pa

8 0
2 years ago
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Solve for x in the equation x squared + 14 x + 17 = negative 96. x = negative 7 plus-or-minus 4 StartRoot 6 EndRoot i x = –7 ± 8
blagie [28]

Answer:

x=-7\pm8i

Step-by-step explanation:

we have

x^{2} +14x+17=-96

Equate to zero

x^{2} +14x+17+96=0

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The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

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

in this problem we have

x^{2} +14x+113=0

so

a=1\\b=14\\c=113

substitute in the formula

x=\frac{-14\pm\sqrt{14^{2}-4(1)(113)}} {2(1)}

x=\frac{-14\pm\sqrt{-256}} {2}

Remember that

i=\sqrt{-1}

so

x=\frac{-14\pm16i} {2}

x=-7\pm8i

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Question 10 Multiple Choice Worth 1 points)
zheka24 [161]

Answer:

Linearly, because the table shows that the sunflowers increased by the same amount each month

Step-by-step explanation:

Given the table

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Note that months change one-by-one (21-1, 3-2=1, 4-3=1).

Also

17.2-15=2.2\ [\text{from month 1 to month 2}]\\ \\19.4-17.2=2.2\ [\text{from month 2 to month 3}]\\ \\21.6-19.4=2.2\ [\text{from month 3 to month 4}]

This means the number of sunflowers increases linearly, because the table shows that the sunflowers increased by the same amount each month

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