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andrew11 [14]
3 years ago
14

A rectangle is to be inscribed in a right triangle having sides of length 6 in, 8 in, and 10 in. Find the dimensions of the rect

angle with greatest area assuming the rectangle is positioned as in Figure 1. Figure1

Mathematics
1 answer:
Helga [31]3 years ago
8 0

Answer:  width = 2.4 in, length = 5

<u>Step-by-step explanation:</u>

The max area of a right triangle is half the area of the original triangle.

Area of the triangle = (6 x 8)/2  = 24

--> area of rectangle = 24 ÷ 2 = 12

Next, let's find the dimensions.

The length is adjacent to the hypotenuse. Since we know the area is half, we should also know that the length will be half of the hypotenuse.

length = 10 ÷ 2 = 5

Use the area formula to find the width:

A = length x width

12 = 5 w

12/5 = w

2.4 = w

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Question: Find The Sum Of The Integers From -6 To 58 Of Ss. O 1,500 O 1,734 O 1,690 O 1,621​
REY [17]

Answer:

The Sum Of The Integers From -6 To 58 is <u>1690.</u>

Step-by-step explanation:

Given,

a=-6

T_n=58

We have to find out the sum of integers from -6 To 58.

Firstly we will find out the total number of terms that is 'n'.

Here a_1=-6\ and\ a_2=-5

\therefore d=a_2-a_1=-5-(-6)=-5+6=1

Now we use the formula of A.P.

T_n=a+(n-1)d

On substituting the values, we get;

58=-6+(n-1)1\\\\n-1=58+6\\\\n-1=64\\\\n=64+1=65

So there are 65 terms in between  -6 To 58.

That means we have to find the sum of 65 terms in between  -6 To 58.

Now we use the formula of Sum of n_terms.

S_n=\frac{n}{2}(2a-(n-1)d)

On substituting the values, we get;

S_{65}=\frac{65}{2}(2\times-6+(65-1)1)\\\\S_{65}=\frac{65}{2}(-12+64)\\\\S_{65}=\frac{65}{2}\times52\\\\S_{65}=65\times26=1690

Hence The Sum Of The Integers From -6 To 58 is <u>1690.</u>

5 0
3 years ago
I need some assistance. Please help!
horsena [70]

Answer: Hope this helps you!

Edit: the 2nd photo is more in detail and should help you better. Good luck!

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
The weight Wkg of a metal bar varies jointly
maria [59]

Answer:

d = \sqrt{\frac{216W}{35L} }

Step-by-step explanation:

Given that W varies jointly as L and d² then the equation relating them is

W = kLd² ← k is the constant of variation

To find k use the condition W = 140 when d = 4 and L = 54, thus

140 = k × 54 × 4² = 864k ( divide both sides by 864 )

\frac{140}{864} = k , that is

k = \frac{35}{216}

W = \frac{35}{216} Ld² ← equation of variation

Multiply both sides by 216

216W = 35Ld² ( divide both sides by 35L )

\frac{216W}{35L} = d² ( take the square root of both sides )

d = \sqrt{\frac{216W}{35L} }

6 0
3 years ago
Given: f(x)=1-x^2,g(x)=1-2x, and h(x)= 1/x^2+1 <br> find f(g(a+1))<br> find g(1/h(z))
AleksAgata [21]

Answer:

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Step-by-step explanation:

7 0
2 years ago
(×+2) +(4×-122)=180 need help with this problem
alina1380 [7]
So first you need to open the brackets, so it would be x+2+4x-122=180. Then we can add the 4x and the other x to make 5x, and then by doing 2-122 we get -120. This gives us the equation 5x-120=180. We then isolate the variable by moving the -120 to the other side of the equation and becoming a positive, so it would look like 5x=180+120. Then, we have 5x=300. 300 divides by 5 is 60 making X=60. Hope this helps!
6 0
3 years ago
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