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IRISSAK [1]
3 years ago
15

. С M 1. AJKME ALKM 2. s S S SSS congruena

Mathematics
2 answers:
My name is Ann [436]3 years ago
8 0
Whatttttttttttttttt.
is this a proof. 5
goldenfox [79]3 years ago
3 0

SSSSSSSSSSSSSSSSSSSSSSSSSSS COOL

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A rectangular solid with a square base has a volume of 4096 cubic inches - determine the dimensions that yield the minimum surfa
Bumek [7]

Answer:

side of base, a = 10.1 inches, height, h = 40.1 inches

Step-by-step explanation:

Volume of rectangular solid, V = 4096 cubic inches

Let the side of base is a and the height is h.

V = a^2h\\\\4096 = a^2 h ..... (1)

surface area of the solid  

S = 2a^2 + 4 ah \\\\S = 2a^2 + 4 \times a \times \frac{4096}{a^2}    from (1)\\\S = 2a^2 + \frac{4096}{a}\\\\\frac{dS}{da}= 4 a - \frac{4096}{a^2}\\\\4 a - \frac{4096}{a^2} = 0 \\\\a^3 = 1024\\\\a =10.1 inches

So, h = 40.2 inches

5 0
3 years ago
I will mark brainliest!
Harrizon [31]

Answer:

a + 56.75 = 76.75

Step-by-step explanation:

that is because ,money raised on first day(a) + money raised on second day = 76.75

that gives us,  a = $20

3 0
3 years ago
Read 2 more answers
A eagle has landed in a tree 50 feet above sea level. Directly below the eagle, a seagull is flying 17 feet above sea level. Dir
Luba_88 [7]

Answer:

As attached

Step-by-step explanation:

6 0
3 years ago
please help me, Prove a quadrilateral with vertices G(1,-1), H(5,1), I(4,3) and J(0,1) is a rectangle using the parallelogram me
mestny [16]

Answer:

Step-by-step explanation:

We are given the coordinates of a quadrilateral that is G(1,-1), H(5,1), I(4,3) and J(0,1).

Now, before proving that this quadrilateral is a rectangle, we will prove that it is a parallelogram. For this, we will prove that the mid points of the diagonals of the quadrilateral are  equal, thus

Join JH and GI such that they form the diagonals of the quadrilateral.Now,

JH=\sqrt{(5-0)^{2}+(1-1)^{2}}=5 and

GI=\sqrt{(4-1)^{2}+(3+1)^{2}}=5

Now, mid point of JH=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})

=(\frac{5+0}{2},\frac{1+1}{2})=(\frac{5}{2},1)

Mid point of GI=(\frac{5}{2},1)

Since, mid point point of JH and GI are equal, thus GHIJ is a parallelogram.

Now, to prove that it is a rectangle, it is sufficient to prove that it has a right angle by using the Pythagoras theorem.

Thus, From ΔGIJ, we have

(GI)^{2}=(IJ)^{2}+(JG)^{2}                             (1)

Now, JI=\sqrt{(4-0)^{2}+(3-1)^{2}}=\sqrt{20} and GJ=\sqrt{(0-1)^{2}+(1+1)^{2}}=\sqrt{5}

Substituting these values in (1), we get

5^{2}=(\sqrt{20})^{2}+(\sqrt{5})^{2} }

25=20+5

25=25

Thus, GIJ is a right angles triangle.

Hence, GHIJ is a rectangle.

Also, The diagonals GI=\sqrt{(4-1)^{2}+(3+1)^{2}}=5  and HJ=\sqrt{(0-5)^2+(1-1)^2}=5 are equal, thus, GHIJ is a rectangle.

6 0
3 years ago
During the 2007 baseball season, Wade was up to bat 10 more times than
bearhunter [10]

Answer:

The number of times Ellis get to bat  is  558.

Step-by-step explanation:

Here, let us assume the number of times Ellis bat = m

So, the number of times Dwight bat =  17 times fewer than Ellis

                                                            = m - 17

Also, number of times Wade got to bat  = 10 more times than Dwight

= ( m- 17 )+ 10

Total number of bat times = 1650

So, the number of times ( Wade + Dwight + Ellis) bat together = 1650

⇒   ( m- 17 )+ 10 + (m - 17) + m =  1650

or, 3 m - 34 + 10 =  1650

or, 3  m = 1674

⇒ m = 1674/3 = 558 , or m = 558

Hence, the number of times Ellis get to bat = m = 558.

3 0
3 years ago
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