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3 years ago

Good at math & need points? help!!

Mathematics

1 answer:

sergeinik [125]3 years ago

4 0

Answer:

26 miles

Step-by-step explanation:

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alisha [4.7K]

Diagonal of the parallelogram divides the parallelogram in to two equal areas.

So area of parallelogram = 2(area of triangle)

According to the given diagram,

AB= 8, AD = 5 and BD = 11

So according to the Heron's formula,

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}\\\\where,\\ s =\frac{a + b + c }{2}$

and a, b and c are the three sides of the triangle

Area of triangle ABD = $s=\frac{8 + 5 + 11}{2} \\\\s = \frac{24}{2}\\\\s = 12Area of triangle ABD = \sqrt{12(12 - 8) (12 - 5) (12 - 11)}\\\\Area of triangle ABD = \sqrt{12(4) (7) (1)}\\\\Area of triangle ABD = \sqrt{336}\\\\Area of triangle ABD = 18.33$

So, area of parallelogram ABCD = 2(area of triangle ABD)

area of parallelogram ABCD = 2 (18.33)

area of parallelogram ABCD = 36.66

area of parallelogram ABCD = 36.7 sq. units

6 0

3 years ago

Read 2 more answers

Evaluate the line integral c y3 ds,<br> c.x = t3, y = t, 0 â‰¤ t â‰¤ 3

Temka [501]

$\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\int_{t=0}^{t=3}t^3\sqrt{1^2+(3t^2)^2}\,\mathrm dt=\int_0^3t^3\sqrt{1+9t^4}\,\mathrm dt$

Take $u=1+9t^4$ so that $\mathrm du=36t^3\,\mathrm dt$ . Then

$\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\frac1{36}\int_{u=1}^{u=730}\sqrt u\,\mathrm du=\frac{730^{3/2}-1}{54}$

7 0

3 years ago

three children each ate 1/4 of a candy bar.Whitch expression shows how you would solve for the totalamount of candy bar that was

Art [367]

A because it shows you the full amount

8 0

3 years ago

Read 2 more answers

What is the value of 7 in the number 21,780

Georgia [21]

I need more information

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2 years ago

HELPPPPPPPP NOWWWW! WILL MARK BRAINLIESTTTTT!

OLEGan [10]

1. slope of the given line = 1/5
the equation is
(y-2)/(x-(-2))=1/5
x+2=5y-10
x-5y+12=0

2. slope of the given line = -1/6
the equation is
(y-9)/x=6
y-9=6x
y=6x+9

5 0

3 years ago