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3 years ago

Solve this inequality: 3b – 7 < 32

Mathematics

2 answers:

forsale [732]3 years ago

5 0

3b - 7 < 32 => 3b < 32 + 7 => 3b < 39 => b < 39÷ 3 => b < 13 Hope it helps !!!

arsen [322]3 years ago

3 0

3b - 7 < 32
3b < 39
b < 13

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Find the dimensions of a rectangle with perimeter 76 m whose area is as large as possible. m (smaller value) m (larger value)

Sladkaya [172]

Answer:

19 m

Step-by-step explanation:

Perimeter = 2(length + width) ; P = 2(l+w) - - (1)

Area = Length * width ; A = l*w - - - (2)

76 = 2(l+w)

76/2 = l+w

l+w = 38

l = 38 - w

Put l = 38 - w in (1)

A = (38-w)*w

A = 38w - w²

At maximum point:

dA/dw = 0

dA/dw = 38 - 2w

38 - 2w = 0

38 = 2w

w = 38/2

w = 19

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A trapezoid has bases that measure 10 cm and 6 cm. The height of the figure is 15 cm. What is the area of th

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Step-by-step explanation:

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3 years ago

There is 46% hard working Elfs in the North Pole. there is 300 elfs in north pole. How many elfs are hard working

sveticcg [70]

Answer:

138 hard working elfs

Step-by-step explanation:

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2 years ago

Read 2 more answers

Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE

inn [45]

Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to $20\ units$

Part 3) The area is equal to $25\ units^{2}$

Step-by-step explanation:

we have

$A(5,0), B(2,4), C(-2,1),D(1,-3)$

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$A(5,0),B(2,4)$

substitute in the formula

$d=\sqrt{(4-0)^{2}+(2-5)^{2}}$

$d=\sqrt{(4)^{2}+(-3)^{2}}$

$d=\sqrt{25}$

$AB=5\ units$

Find the distance BC

$B(2,4), C(-2,1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$BC=5\ units$

Find the distance CD

$C(-2,1),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-1)^{2}+(1+2)^{2}}$

$d=\sqrt{(-4)^{2}+(3)^{2}}$

$d=\sqrt{25}$

$CD=5\ units$

Find the distance AD

$A(5,0),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-0)^{2}+(1-5)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$AD=5\ units$

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

$B(2,4),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-4)^{2}+(1-2)^{2}}$

$d=\sqrt{(-7)^{2}+(-1)^{2}}$

$BD=\sqrt{50}\ units$

Verify if the polygon is a square

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

$BD^{2}=AD^{2}+AB^{2}$

substitute

$(\sqrt{50})^{2}=5^{2}+5^{2}$

$50=50$ -----> is true

The triangle BDA is a right triangle

therefore

The polygon is a square

Find the Area of the polygon

The area of a square is equal to

$A=b^{2}$

we have

$b=5\ units$

$A=5^{2}=25\ units^{2}$

Find the perimeter of the polygon

The perimeter of a square is equal to

$P=4b$

we have

$b=5\ units$

$P=4(5)=20\ units$

6 0

3 years ago

the perimeter of a square is equal to the perimeter of an equilateral triangle the length of a side of the square is given by th

diamong [38]

Answer:

x = 3

Step-by-step explanation:

We are told that the length of a side of the square is x.

Now, perimeter of square = 4 × side length

Perimeter of square = 4x

Also,we are told that the triangle is equilateral and a side is (x + 1).

Thus, perimeter of triangle = 3(x + 1)

Since we are told that the perimeter of the square is equal to that of the equilateral triangle. Thus;

4x = 3(x + 1)

4x = 3x + 3

Subtract 3x from both sides to get;

4x - 3x = 3

x = 3

5 0

3 years ago