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marta [7]
3 years ago
11

Given: ∆ABC, AB = BC, m∠1<90°. Perimeter of ∆ABC = 25. Difference between two sides is 4. Find: AB, BC, AC. Can someone help

me !!!!
Mathematics
1 answer:
goldfiish [28.3K]3 years ago
8 0

Answer:

AB = 7, BC = 7 and AC = 11

Step-by-step explanation:

Let AB = BC = a and CA = b

Now, given that the perimeter of the triangle is 25.

So, 2a + b = 25 ............ (1)

And 4 is the difference between the two unequal sides.

Hence, b - a = 4 .............. (2)  

⇒ 2b - 2a = 8 ........... (3)

Now, solving equations (1) and (3) we get, 3b = 25 + 8 = 33

⇒ b = 11

Hence, from equation (2) we get a = b - 4 = 7

Therefore, AB = 7, BC = 7 and AC = 11 (Answer)

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Math<br><br><br><br> pls help!!<br><br><br><br><br><br> answers?
statuscvo [17]

Answer: Choice B) Infinitely many solutions

  • one solution: x = 8, y = -7/2, z = 0
  • another solution: x = -12, y = 13/2, z = 10

=======================================================

Explanation:

Here's the starting original augmented matrix.

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\-4 & 0 & -8 & -32\\\end{array}\right]

We'll multiply everything in row 3 (abbreviated R3) by the value -1/4 or -0.25, which will make that -4 in the first column turn into a 1.

We use this notation to indicate what's going on: (-1/4)*R3 \to R3

That notation says "multiply everything in R3 by -1/4, then replace the old R3 with the new corresponding values".

So we have this next step:

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\1 & 0 & 2 & 8\\\end{array}\right]\begin{array}{l}  \ \\\ \\(-1/4)*R3 \to R3\\\end{array}

Notice that the new R3 is perfectly identical to R1.

So we can subtract rows R1 and R3, and replace R3 with the result of nothing but 0's

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l}  \ \\\ \\R3-R1 \to R3\\\end{array}

Whenever you get an entire row of 0's, it <u>always</u> means there are infinitely many solutions.

-------------------

Now let's handle the second row. That 5 needs to turn into a 0. We can multiply R1 by 5, and subtract that from R2.

So we need to compute 5*R1-R2 and have that replace R2.

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\0 & 1 & -1 & -7/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l}  \ \\5*R1-R2 \to R2\ \\\ \\\end{array}

Notice that in the third column of R2, we have 9-5*2 = 9-10 = -1. So we have -1 replace the 9. In the fourth column of R2, we have 73/2 - 5*8 = -7/2. So the -7/2 replaces the 73/2.

--------------------

At this point, the augmented matrix is in RREF form. RREF stands for Reduced Row Echelon Form. It seems a bit odd that the "F" of "RREF" stands for "form" even though we say "form" right after "RREF", but I digress.

Because the matrix is in RREF form, this means R1 and R2 lead to these equations:

R1 : 1x+0y+2z = 8\\ R2: 0z+1y-1z = -7/2

which simplify to

R1: x+2z = 8\\R2: y-z = -7/2

Let's get the z terms to each side like so:

x+2z = 8\\x = -2z+8\\\text{ and }\\y-z = -7/2\\y = z-7/2\\

Therefore, all of the solutions are of the form (x,y,z) = (-2z+8, z-7/2, z) where z is any real number.

If z is allowed to be any real number, then we can simply pick any number we want to replace it. We consider z to be the "free variable", in that it's free to be whatever it wants. The values of x and y will depend on what we pick for z.

So the concept of "infinitely many solutions" doesn't exactly mean we can pick just <em>any</em> triple for x,y,z (admittedly it would be nice to randomly pick any 3 numbers off the top of my head and be done right away). Instead, we can pick anything we want for z, and whatever we picked, will directly determine x and y. The x and y are locked into place so to speak.

Let's say we picked z = 0.

That would lead to...

x = -2z+8\\x = -2(0)+8\\x = 8\\\text{ and }\\y = z-7/2\\y = 0-7/2\\y = -7/2\\

So z = 0 would lead to x = 8 and y = -7/2

Rearranging the items in alphabetical order gets us:

x = 8, y = -7/2, z = 0

We have one solution of (x,y,z) = (8, -7/2, 0)

Now let's say we picked z = 10

x = -2z+8\\x = -2(10)+8\\x = -12\\\text{ and }\\y = z-7/2\\y = 10-7/2\\y = 13/2\\

So we have x = -12, y = -13/2, z = 10

Another solution is (x,y,z) = (-12, 13/2, 10)

There's nothing special about z = 0 or z = 10. You can pick any two real numbers you want for z. Just be sure to recalculate the x and y values of course.

To verify each solution, you'll need to plug them back into the original equations formed by the original augmented matrix. After simplifying, you should get the same thing on both sides.

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2 years ago
Determine whether the function f(x) = 3x4 is even or odd.
Lana71 [14]

Answer:

i am 4 th

Step-by-step explanation:

7 0
3 years ago
Find the slope of (4,4) (-6-4) plz need this like in 20 minutes
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Step-by-step explanation:

6 0
3 years ago
At RSM summer camp 100 students had enough food for a 30-days stay. After 2 days 40 new students joined the camp. After that 2 w
omeli [17]

Answer:

"no additional days of food need to be ordered."

Step-by-step explanation:

Suppose initial quantity of food is x units

Assuming each student eats 1 unit of food per day,

x = 30 * 100 = 3000 units of food

In 2 days, food eaten:

100 * 1 * 2 = 200 units

So food left after 2 days is  3000 - 200 = 2800 units

Now, there are 140 students. They stay 2 weeks (14 days) - 2 = 12 more days

So food eaten in 12 days:

140 * 1 * 12 = 1680 units

<u>THus, after total 14 days gone, the amount of food left is:</u>

2800 - 1680 = 1120 units

Half students left, so there are 70 students left for the last 16 days. How much food would they need?

70 * 1 * 16 = 1120 units

And there are exactly 1120 units left. So, no additional days of food need to be ordered.

6 0
3 years ago
70:100 in whole form
taurus [48]
The Answer Would Be 70
6 0
3 years ago
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