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2 years ago

How do I find the median for a congruent triangle

1 answer:

3 0

The length of a median is equal to half the square root of the difference of twice the sum of the squares of the two sides of the triangle that include the vertex the mediam is drawn from and the square of the side of the triangle the median is drawn to.

triangle sides by a, b, c.

ma=122c2+2b2−a2

mb=122c2+2a2−b2

mc=122a2+2b2−c2

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Colin received a bonus check for 2,500$. He is going to deposit the money into his bank account that receives 5.5% simple intere

alex41 [277]

This is the concept of financial mathematics, the simple interest rate is given by:
S.I.=(PRT)/100
where:
P=principle=$2500
R=rate=5.5%
T=time=1 year
Hence;
S.I.=(5.5*2500.1)/100
=$137.5
Thus the total amount he had after 1 year will be:
2500+137.5
=$2,637.5

6 0

3 years ago

If m RDC = 120°, then m DAC=?

monitta

The correct answer to this question is 240

From the image i've seen, the RDC has an angle of 120. So the major arc DC (that passes A and B) has an angle of 240.

So this makes that the measure of the angle of DAC is 240 because of the angle formed by RDC which is 120.

7 0

3 years ago

Read 2 more answers

A piece of cardboard is 13 inches by 26 inches. A square is to be cut from each corner and the sides folded up to make an open-t

Vanyuwa [196]

Answer:

Hence the maximum possible volume will be the 778.53 c.c

Step-by-step explanation:

Given:

A rectangle with 13 x 26 dimensions

And corners are cut to form side squares.

To Find:

Maximum possible volume for box

Solution :

Consider a rectangle of 13 x 26 dimension with and side of square at corner be x.

(Refer the attachment)

Now,

Formulating the volume equation for the box

So corner square sides we are going to fold up which makes height of the box

and remaining part will be length and breadth

As shown in fig,

Length=26-x

breadth=13-x

And height will be x

$V(x)=x*(26-x)*(13-x)$

To get maximum volume differentiate the above equation,

$V(x)=x*(26*13-26*x-13*x+x^2)$

$V(x)=x^3-39x^2+338x\\$

$V'(x)=3x^2-78x+338$

$V''(x)=6x-78$

Now ,Solve the Quadratic Equation to get x values,

$3x^2-78x+338$ =0

x=[-b±(b^2-4ac)^1/2]/2a

x=[78±Sqrt[(78)^2-4*338*3)]/2*3

x=[78±Sqrt(3028)]/6

x=[78±55.027]/6

x=78+55.027/6 or x=78-55.027/6

x=22.17 or x=3.8288

Use these values in 6x-78 to know which value posses the max and min value for the function.

So when x=22.17

6x-78=6*22.17-78

=55.02>0 i.e function will have minimum value .

When x=3.8288

6*3.8288-78

=-55.0272<0 i.e. Function will have maximum value

Now, the function will defines the maximum volume

$V(x)=x^3-39x^2+338x$

$V(x)=3.8288^3-39*(3.82883)^2+338*3.8288$

$V(x)=56.13-571.73+1294.13$

$V(x)=778.53 C.C$

6 0

3 years ago

What is the midpoint between (10,4) and (-6,-4)

yanalaym [24]

Answer:

(2,0)

Step-by-step explanation:

Take the average of the two x-coordinates: (10-6)/2 = 2

Take the average of the two y-coordinates: (4+(-4))/2 = 0

Midpoint: (2,0)

4 0

2 years ago

Read 2 more answers

Forgot to do this over winter break. Its due tomorrow and I haven't used my brain for 2 weeks. HELPPP (may have to zoom in)

ANEK [815]

1. 60,30,90 right triangle. y will be hypotenuse/2, x will be
hypotenuse*sqrt(3)/2. So x = 16*sqrt(3)/2 = 8*sqrt(3), approximately 13.85640646
y = 16/2 = 8
2. 45,45,90 right triangle (2 legs are equal length and you have a right angle).
X and Y will be the same length and that will be hypotenuse * sqrt(2)/2. So
x = y = 8*sqrt(2) * sqrt(2)/2 = 8*2/2 = 8
3. Just a right triangle with both legs of known length. Use the Pythagorean theorem
x = sqrt(12^2 + 5^2) = sqrt(144 + 25) = sqrt(169) = 13
4. Another right triangle with 1 leg and the hypotenuse known. Pythagorean theorem again.
y = sqrt(1000^2 - 600^2) = sqrt(1000000 - 360000) = sqrt(640000) = 800  5. A 45,45,90 right triangle. One leg known. The other leg will have the same length as the known leg and the hypotenuse can be discovered with the Pythagorean theorem.  x = 6. y = sqrt(6^2 + 6^2) = sqrt(36+36) = sqrt(72) = sqrt(2 * 36) = 6*sqrt(2), approximately 8.485281374
6. Another 45,45,90 triangle with the hypotenuse known. Both unknown legs will have the same length. And Pythagorean theorem will be helpful.
x = y.
12^2 = x^2 + y^2
12^2 = x^2 + x^2
12^2 = 2x^2
144 = 2x^2
72 = x^2
sqrt(72) = x
6*sqrt(2) = x
x is approximately 8.485281374
7. A 30,60,90 right triangle with the short leg known. The hypotenuse will be twice the length of the short leg and the remaining leg can be determined using the Pythagorean theorem.
y = 11*2 = 22.
x = sqrt(22^2 - 11^2) = sqrt(484 - 121) = sqrt(363) = sqrt(121 * 3) = 11*sqrt(3). Approximately 19.05255888
8. A 30,60,90 right triangle with long leg known. Can either have fact that in that triangle, the legs have the ratio of 1:sqrt(3):2, or you can use the Pythagorean theorem. In this case, I'll use the 1:2 ratio between the unknown leg and the hypotenuse along with the Pythagorean theorem.
x = 2y
y^2 = x^2 - (22.5*sqrt(3))^2
y^2 = (2y)^2 - (22.5*sqrt(3))^2
y^2 = 4y^2 - 1518.75
-3y^2 = - 1518.75
y^2 = 506.25 = 2025/4
y = sqrt(2025/4) = sqrt(2025)/sqrt(4) = 45/2
Therefore:
y = 22.5
x = 2*y = 2*22.5 = 45
9. Just a generic right triangle with 2 known legs. Use the Pythagorean theorem.
x = sqrt(16^2 + 30^2) = sqrt(256 + 900) = sqrt(1156) = 34
10. Another right triangle, another use of the Pythagorean theorem.
x = sqrt(50^2 - 14^2) = sqrt(2500 - 196) = sqrt(2304) = 48

8 0

3 years ago