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vichka [17]
3 years ago
8

Find the area of the quadrilateral in the figure.

Mathematics
1 answer:
kirill [66]3 years ago
4 0
<h3>Answer:</h3>

C. 19.64

<h3>Explanation:</h3>

The triangle at upper right is a 3-4-5 right triangle, so has an area that is half the product of the leg lengths:

... upper right area = (1/2)(3 units)(4 units) = 6 units²

The triangle at lower left is an isosceles triangle with base length 5 and side length 6. The altitude to the side of length 5 is a bisector of that side and forms right angles at the point of intersection. Hence we can use the Pythagorean theorem to find the triangle's altitude:

... lower left triangle altitude = √(6² - 2.5²) = √29.75 ≈ 5.45436

Then the area of the lower left triangle is half the product of this altitude and the base length of 5 units:

... lower left area = (1/2)(5.45436 units)(5 units) ≈ 13.6359 units²

The quadrilateral's area is the sum of the areas of these triangles, so is ...

... quadrilateral area = upper right area + lower left area

... = 6 units² + 13.6359 units²

... = 19.6359 units² ≈ 19.64 units²

_____

Confirmed by my geometry program as shown in the attachment.

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in a first aid kit the ratio of large bandages to small bandages is 5 to 2. Based on this ratio, how many large bandages are in
Radda [10]
The best way to answer this question is to set up the first sentence as a ratio of large bandages to total bandages. You would write 5 large bandages to 7 total bandages to 7 total bandages.  Then you would make this equivalent to x number of large bandages to 60 total.  It would look like this 5:7 = x:60.  You could use cross products to multiply 60 by 5 to get 300.  7 times x also should equal 300.  Unfortunately, this example will not leave us with a whole number of bandages, but 300/7 is a repeating decimal or a mixed number (42 6/7 large bandages). 
5 0
3 years ago
Plz with steps .. it's very hard can anyone plz
liubo4ka [24]

Answer:

Step-by-step explanation:

\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a}  \qquad  f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a}  \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}}  }{\dfrac{1}{2\sqrt{x}} }\\\\

\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}  =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\

7 0
3 years ago
1. What is the measure in radians for the central angle of a circle whose radius is 6 cm and intercept arc length is 5.4 cm?
fgiga [73]
Hello!

1.
find circumferece
c=2pir
c=2*6*pi=12pi
intercept is 5.4

2pi radians=all

part/whole=part/whole
5.4/12pi=x/2pi
times both sides by 12pi
5.4=6x
divide both sides by 6
0.9=x
answer is 0.9 radians



2. 
assuming 9pi/5 radians

find circumference
c=2pir
c=2*26.9*pi
c=53.8pi

arc/circumference=(9pi/5)/2pi

x/(53.8pi)=(9pi/5)/(2pi)
x/(53.8pi)=18/5
times both sides by 53.8pi
x=608.46266 m
about x=608.46 m


Hope this Helps! Have A Wonderful Day! :) 
5 0
3 years ago
Jenny makes quilts. she can make 7 quilts with 21 yards of material. how many yards of material would be required to make 12 qui
kozerog [31]
3 yards per quilt

36 for 12
8 0
3 years ago
Read 2 more answers
Does (-24,-12) and (-15,-9) line pass through point (18,2)?
Dmitrij [34]

Find the slope using the formula [ y2-y1/x2-x1 ].

-9-(-12)/-15-(-24)

3/9

1/3

Find the y-intercept using the formula [ y = mx + b ].

y = 1/3x + b

-12 = 1/3(-24) + b

-12 = -8 + b

-4 = b

Use (18, 2) to see if the line passes through it

y = 1/3x - 4

2 = 1/3(18) - 4

2 = 6 - 4

2 = 2 TRUE

Thus, the line does pass through (18, 2).

Best of Luck!

4 0
2 years ago
Read 2 more answers
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