All categories

3 years ago

Adam has sticks of the following lengths: (3x+4) (2x+10) (x+12)cm

2 answers:

3 0

Option I
(3x+4)cm =(2x+10)cm
3x-2x=6
x=6
Option II
(3x+4)= (x+12)
3x-x=12-4
2x=8
x=4
Option III
(2x+10)= (x+12)
2x-x=12-10
x=2

yaroslaw [1]3 years ago

3 0

Isosceles triangles, are triangles where two of the lengths are the same so we can say that:
1. (3x+4)=(2x+10)
2. (3x+4)=(x+12)
3. (2x+10)=(x+12)
One of these must be true. Solve as simultaneous equations.
1. x=6
2. x=4
3. x=2

You might be interested in

Is y = 3(x - 2) linear or nonlinear

mixas84 [53]

Answer:

The equation is Linear

Step-by-step explanation:

6 0

2 years ago

- The four-digit number 1210 has an interesting property.

vampirchik [111]

Answer:

ABCDEFG = 3211000

Step-by-step explanation:

A counts the number of zeroes, and there are 3 zeroes (E, F, G), so A = 3.B counts the number of ones, and there are 2 ones (C, D), so B = 2.C counts the number of twos, and there is 1 two (B), so C = 1.D counts the number of threes, and there is 1 three (A), so D = 1.There is no four, five, six in, so E, F, G are all zeroes.

5 0

2 years ago

There are 32 sixth graders and 44 seventh graders in the video gaming club. Which ratio is the greatest?

katrin [286]

I’m gonna say D hope this helps

3 0

3 years ago

Evaluate 5y+x2÷2 when x=4 and y=−3. When x=4 and y=−3, 5y+x2÷2=

Alex Ar [27]

The answer is 5 x -3 + 4 x 2 / 2 = -15 + 8 / 2 = -11

6 0

3 years ago

This is Matrix for pre calc

gavmur [86]

The given system of equations in augmented matrix form is

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]$

If you need to solve this, first get the matrix in RREF:

Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]$

Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]$

Add 164(row 3) to -91(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]$

Multiply row 4 by 1/13080:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]$

Add -153(row 4) to row 3:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]$

Multiply row 3 by -1/91:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add 6(row 3) and -8(row 4) to row 2:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 2 by 1/5:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

$\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 1 by 1/3:

$\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

So the solution to this system is $(w,x,y,z)=(1,-2,4,-3)$ .

6 0

3 years ago