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kodGreya [7K]
3 years ago
8

What is the area of the figure below? 7.5m^2 15m^2 21.25m^2 42.5m^2

Mathematics
2 answers:
Novay_Z [31]3 years ago
7 0
The answer is c!!! hope this helps!!!
Olenka [21]3 years ago
6 0

Answer:

Third option is correct. The area of figure is 21.25m^2.

Step-by-step explanation:

The area of a kite is

A=\frac{1}{2}(d_1\times d_2)

Both diagonals are perpendicular. Longer diagonal bisects the smaller diagonal.

In triangle ABO,

\tan45^{\circ}=\frac{BO}{AO}

1=\frac{BO}{2.5}

2.5=BO

The length of BD is

BD=2\times BO=2\times 2.5=5

The length of AC is

6+2.5=8.5

The area of kite ABCD is

A=\frac{1}{2}(AC\times BD)

A=\frac{1}{2}(8.5\times 5)

A=21.25

Therefore third option is correct. The area of figure is 21.25m^2.

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3x - 5y = 15<br> x - 4y = 12<br> Substitution method
Scorpion4ik [409]

Answer:

x=0,y=-3

Step-by-step explanation:

3x-5y=15 - equation(1)

x-4y=12 => x=12+4y -equation (2)

Replace equation (2) in equation (1)

3(12+4y)-5y=15

36+12y-5y=15

7y=15-36

7y= -21

y = -21/7

y=-3

If y=-3,then replace in equation (1) ; 3x -5(-3)=15

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x=0

4 0
3 years ago
Which is the correct answer​
Sav [38]

Answer:

Option 3

Step-by-step explanation:

Using the order of operations, you will find out that 7 x 10^3 = 7,000, and 3 x 10^2 = 300. Dividing 7,000 by 300 will get you 23.333~, so 300 does go into 7,000 a little more than 20 times.

8 0
2 years ago
An airliner maintaining a constant elevation of 2 miles passes over an airport at noon traveling 500 mi/hr due west. At 1:00 PM,
butalik [34]

Answer:

\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}

Step-by-step explanation:

Let suppose that airliners travel at constant speed. The equations for travelled distance of each airplane with respect to origin are respectively:

First airplane

r_{A} = 500\,\frac{mi}{h}\cdot t\\r_{B} = 550\,\frac{mi}{h}\cdot t

Where t is the time measured in hours.

Since north and west are perpendicular to each other, the staight distance between airliners can modelled by means of the Pythagorean Theorem:

s=\sqrt{r_{A}^{2}+r_{B}^{2}}

Rate of change of such distance can be found by the deriving the expression in terms of time:

\frac{ds}{dt}=\frac{r_{A}\cdot \frac{dr_{A}}{dt}+r_{B}\cdot \frac{dr_{B}}{dt}}{\sqrt{r_{A}^{2}+r_{B}^{2}} }

Where \frac{dr_{A}}{dt} = 500\,\frac{mi}{h} and \frac{dr_{B}}{dt} = 550\,\frac{mi}{h}, respectively. Distances of each airliner at 2:30 PM are:

r_{A}= (500\,\frac{mi}{h})\cdot (1.5\,h)\\r_{A} = 750\,mi

r_{B}=(550\,\frac{mi}{h} )\cdot (1.5\,h)\\r_{B} = 825\,mi

The rate of change is:

\frac{ds}{dt}=\frac{(750\,mi)\cdot (500\,\frac{mi}{h} )+(825\,mi)\cdot(550\,\frac{mi}{h})}{\sqrt{(750\,mi)^{2}+(825\,mi)^{2}} }

\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}

6 0
3 years ago
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uranmaximum [27]
It is a translation
5 0
3 years ago
Bee + ant= 10, ant - bee = 8, ant × bee= ?
Alona [7]

The Correct Answer Is...

<em><u>9.</u></em>

Any Questions? Comment Below!

<em><u>-AnonymousGiantsFan</u></em>

7 0
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