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baherus [9]
3 years ago
12

The ratio of the measures of the sides of a triangle is 7:9:12, and it's perimeter is 84 inches. Find the measures of each side

of the triangle.
Mathematics
1 answer:
Ede4ka [16]3 years ago
6 0

Answer:

Step-by-step explanation:

The lengths of the sides are 7x, 9x, and 12x.

7x + 9x + 12x = 84

28x = 84

x = 3 in

Sides: 21 in, 27 in, 36 in

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1575 would be about correct!

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1/(x-2)+1/(x+3)=1/5 solve the problem for me​
Gennadij [26K]

════════ ∘◦❁◦∘ ════════

<h3>Final value : x² -9x - 11</h3>

════════════════════

<h3>Step by step </h3>

\frac{1}{(x - 2)}  +  \frac{1}{(x + 3)}  =   \frac{1}{5}

\frac{(x + 3) + (x - 2)}{(x - 2)(x + 3)}  =  \frac{1}{5}

\frac{2x + 1}{ {x}^{2} + 3x - 2x - 6 }  =  \frac{1}{5}

\frac{2x + 1}{ {x}^{2}  + x - 6}  =  \frac{1}{5}

2x + 1 = ⅕ × (x² + x - 6)

2x + 1 = ⅕x² + ⅕x - 6/5

1/5x² +1/5x - 2x - 6/5 - 1 = 0

1/5x² -9/5x - 11/5 = 0

x² - 9x - 11 = 0. #times by 5

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#Give me brainliest pls im tired typing all of this

8 0
3 years ago
Show work please.<br><br> solve system of equations using matrices.
nadya68 [22]

Answer:

(t, t -1, t)

Step-by-step explanation:

You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called.  I think it refers to infinite many solutions).  Here's how it works:

Set up your matrix:

\left[\begin{array}{ccc}1&-2&1\\2&-1&-1\\\end{array}\right] \left[\begin{array}{ccc}2\\1\\\end{array}\right]

You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left.  Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0.  Multiplying in a -2 to the top row gives you:

\left[\begin{array}{ccc}-2&4&-2\\2&-1&-1\\\end{array}\right]\left[\begin{array}{ccc}-4\\1\\\end{array}\right]

Then add, keeping the first row the same and changing the second to reflect the addition:

\left[\begin{array}{ccc}-2&4&-2\\0&3&-3\\\end{array}\right] \left[\begin{array}{ccc}-4\\-3\\\end{array}\right]

The second equation is this now:

3y - 3z = -3.  Solving for y gives you y = z - 1.  Let's let z = t (some random real number that will make the system true.  Any number will work.  I'll show you at the end.  Just bear with me...)

lf z = t, and if y = z - 1, then y = t - 1.  So far we have that y = t - 1 and z = t.  Now we solve for x:

From the first equation in the original system,

x - 2y + z = 2.  Subbing in t - 1 for y and t for z:

x - 2(t - 1) + t = 2.  Simplify to get

x - 2t + 2 + t = 2  and  x - t = 0, and x = t.  So the solution set is (t, t - 1, t).  Picking a random value for t of, let's say 2, sub that in and make sure it works.  If:

x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2.    Check it:  2 - 4 + 4 = 2 and 2 = 2.  You could pick any value for t and it will work.

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Add. 3/10+(−7/20) Enter your answer
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Make the denominator the same, so multiple 3/10 by 2, which produces 6/20. Take 6/20 and subtract 7/20 because adding a negative number is the same as subtracting a positive number. the Answer is -1/20.
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