All categories

3 years ago

Could someone confirm if I was correct or not?

Mathematics

2 answers:

Andrei [34K]3 years ago

7 0

Correct! Excellent Job

larisa86 [58]3 years ago

7 0

I think sure it’s correct!

You might be interested in

Which of the following best describes the slope of the line below?

cluponka [151]

Answer:

That slope is undefined

Step-by-step explanation:

x doesn't have an existing value, therefore it is undefined.

4 0

3 years ago

Hi! I need help asap!!!! Can someone help me on #32!! I need an explanation fast!!

Nutka1998 [239]

5 degrees at 10 pm because it started at 3 then went to 14 which is 17 then went down 12 degrees leaving you with 5 degrees. hope you understand this

4 0

3 years ago

Determine whether the given lengths can be sides of a right triangle. Which of the following are true statements? The lengths 14

devlian [24]

Answer:

The lengths 14, 24, and 26 cannot be the sides of a right triangle. The lengths 30, 72, 78 can be the sides of a right triangle.

Step-by-step explanation:

To prove that the lengths 14, 24, and 26 cannot be the sides of a right triangle:

a=14, b=24, c=26

Pythagoreon theorem: a^2+b^2=c^2

substitute values in: 14^2+24^2=26^2

simplify: 196+576=676

simplify again: 772=676, which is not true

This proves that the lengths 14, 24, and 26 cannot be the sides of a right triangle.

To prove that the lengths 30, 72, and 78 can be the sides of a right triangle:

a=30, b=72, c=78

Pythagoreon theorem: a^2+b^2=c^2

substitute values in: 30^2+72^2=78^2

simplify: 900+5184=6084

simplify again: 6084=6084, which is true

This proves that the lengths 30, 72, and 78 can be the sides of a right triangle.

5 0

3 years ago

Read 2 more answers

If 1 pint is equal to 2 cups, then 5 pints would equal how many cups

zmey [24]

If 1 pint is equal to 2 cups then 5 pints equals 10 cups

5 times 2 equals 10.

4 0

3 years ago

Read 2 more answers

Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

timurjin [86]

Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

6 0

4 years ago