2(m-3)+7<21 PLEASE HELP!!!

Mathematics

1 answer:

Contact [7]2 years ago

4 0

Simplify. Distribute 2 to all terms within the parenthesis

2(m -3) = 2m - 6

Simplify. Combine like terms

2m - 6 + 7 < 21

2m + 1 < 21

Isolate the m. Subtract 1 from both sides

2m + 1 (-1) < 21 (-1)

2m < 20

Isolate the m. Divide 2 from both sides

2m/2 < 20/2

m < 20/2

m < 10

m < 10 is your answer

hope this helps

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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: