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Tcecarenko [31]
3 years ago
14

What was 50.00 worth in 1900?

Mathematics
1 answer:
UNO [17]3 years ago
6 0
$50 in the year 1900 would be worth about $2 today. 
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Segment EF is a mid segment of triangle ABC. If EF = 15 cm, what is the measure of segment BC?
vaieri [72.5K]

Answer:

30

Step-by-step explanation:

6 0
3 years ago
Determine the value of s, the arc length (measured in inches) cut off in a circle with a radius of 3.4 inches by an angle with a
AleksandrR [38]

Answer:

Arc length s = 15.3 inches

Step-by-step explanation:

We know that length of an arc that subtends an angle θ radians at the center is given by

s=r\times \Theta

where

r is the radius of the circle

θ  is the angle subtended in radians

Thus applying values we get s=3.4\times \ 4.5radians

thus s = 15.3 inches

4 0
3 years ago
Find the perimeter of the shaded region
Bond [772]

Answer:

13 u²

Step-by-step explanation:

8 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
A given line has a slope of - ⅚.  (ie, m = -⅚)
Nastasia [14]

Answer:

Step-by-step explanation:

The slope of the line parallel to that line is -5/6 because parallel lines have the same slope.

8 0
3 years ago
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