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pychu [463]
3 years ago
15

Solve for the red variable

Mathematics
1 answer:
pishuonlain [190]3 years ago
3 0

Answer:

1/3 Bh = V

Step-by-step explanation:

B = 3 V / h

Multiply each side by h

Bh = 3 V / h *h

Bh = 3V

Divide each side by 3

1/3 Bh = 3V/3

1/3 Bh = V

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So, 
3 - (-6) / 2 - 2 = 9 / 0 for the first two points. The slope of the line is undefined because it is a vertical line. 
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3 years ago
the perimeter of a square can be found using P = n + n + n + n, where n is the length of one side of the square. Which equation
AlexFokin [52]

Answer:

P = 4n

Step-by-step explanation:

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Solve the inequality for w
kobusy [5.1K]

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w ≥ 23

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3 0
4 years ago
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
melamori03 [73]

Answer:

6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base b and height h is given by A=\frac{1}{2}bh. Therefore, the area of this right triangle is:

A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}

The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

A=\sqrt{s(s-a)(s-b)(s-c)}, where a, b, and c are three sides of the triangle and s is the semi-perimeter (s=\frac{a+b+c}{2}).

The semi-perimeter, s, is:

s=\frac{5+5+4}{2}=\frac{14}{2}=7

Therefore, the area of the isosceles triangle is:

A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}

Thus, the area of the quadrilateral is:

6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}

4 0
3 years ago
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