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liberstina [14]
3 years ago
7

Someone please help me this will be a lot of points and I will follow u and Mark u the brainiest

Mathematics
1 answer:
tamaranim1 [39]3 years ago
8 0

Answer:

C=2v

Step-by-step explanation:

because 0*2=0 1*2=2 2*2=4 3*2=6 4*2=8 5*2=10 so C=2v

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Question 17
SOVA2 [1]

Answer:

1.8 × 10^10

Step-by-step explanation:

(7.2 × 10 = 72)

(72) ÷ (4 × 10^-9) = 1.8 × 10^10

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2 years ago
16y=14+9y please answer 23 points
NeTakaya

Answer:

<h2>y = 2</h2>

Step-by-step explanation:

16y=14+9y\qquad\text{subtract}\ 9y\ \text{from both sides}\\\\16y-9y=14+9y-9y\\\\7y=14\qquad\text{divide both sides by 7}\\\\\dfrac{7y}{7}=\dfrac{14}{7}\\\\y=2

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3 years ago
What is the midpoint of the line segment with endpoints?
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Answer:

B. (1,2)

Step-by-step explanation:

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6 0
3 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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