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gladu [14]
3 years ago
8

Pythagoras question plz help

Mathematics
1 answer:
motikmotik3 years ago
6 0
From the two right triangles, you can write the following equations using the Pythagorean theorem. Let's call that shared leg in the middle "y"

y^2 + b^2 = a^2
y^2 + c^2 = x^2


y^2 + b^2 = a^2
re-write this to get "y" alone for substitution.
y^2 = a^2 - b^2
substitute (a^2 - b^2) for y^2 in the other equation. y^2 + c^2 = x^2

a^2 - b^2 + c^2 = x^2

Now put in the values given for a,b,c to solve for x

(7.1)^2 - (5.6)^2 + (5.7)^2 = x^2
51.54 = x^2
square root
7.2 = x


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Solve the system of equations by finding the reduced row echelon form of the augmented matrix. Write the solutions for x and y i
Gwar [14]

Answer:

The solutions for the given system of equations are:

\left \{ {{x=-5} \atop {y=12-4z}} \right.

Step-by-step explanation:

Given the equation system:

\left \{ {{3x+y+4z=-3} \atop {-x+y+4z=17}} \right.

We obtain the following matrix:

\left[\begin{array}{cccc}3&1&4&-3\\-1&1&4&17\end{array}\right]

<u>Step 1:</u> Multiply the fisrt row by 1/3.

\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\-1&1&4&17\end{array}\right]

<u>Step 2:</u> Sum the first row and the second row.

\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&\frac{4}{3} &\frac{16}{3}&16\end{array}\right]

<u>Step 3:</u> Multiply the second row by 3/4.

\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&1 &4&12\end{array}\right]

<u>Step 4:</u> Multiply the second row by -1/3 and sum the the first row.

\left[\begin{array}{cccc}1&0 &0&-5\\0&1 &4&12\end{array}\right]

The result of the reduced matrix is:

\left \{ {{x=-5} \atop {y+4z=12}} \right.

This is equal to:

\left \{ {{x=-5} \atop {y=12-4z}} \right.

These are the solutions for the system of equations in terms of z, where z can be any number.

6 0
3 years ago
Can somebody please help me with this pleaseee
Semmy [17]

Answer: -2, 2

Step-by-step explanation:

0 = 5x^2 -20

0 = 5(x^2 -4)

0 = 5(x - 2)(x + 2)

0 = x - 2

2 = x

0 = x + 2

-2 = x

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