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3 years ago

Select the values that make the inequality -n ≤ 6 true.

Mathematics

1 answer:

N76 [4]3 years ago

7 0

Answer:

-6 -5.9 -5

-1 0 1

5 5.9 6

6.1 7 11

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GaryK [48]

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Step-by-step explanation:

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3 years ago

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inysia [295]

Answer:

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3 years ago

6x – 3y = 8

Nat2105 [25]

Answer:

Step-by-step explanation:

<em>The linear equation where:</em>

$\large \boldsymbol {} \sf y=\underbrace{m}_{slope }x \ +\underbrace{b} _{y -intersept}$

Solution :

$\displaystyle \sf \#13. \\\\ 6x-3y=8 \\\\\ -3y=8-6x \\\\y=-\frac{8-6x}{3} \\\\ \boxed{\sf y=2x-2\frac{1}{3} }} \\\\ slope =2 \\\\ y-intersept =-2\frac{1}{3} \\\\-------------$

$\sf \#14 . \\\\\\ 7x=5y+2 \\\\5y=7x-2 \\\\y =\dfrac{7x-2}{5} \\\\ \boxed{\sf y=1,4x-0,4} \\\\slope = 1,4 \\\\y-intersept =-0,4 \\\\---------------$

$\dispalystye \sf \#15. \\\\ -6y+4x=8 \ |\div2 \\\\-3y+2x=4 \\\\ y=-\dfrac{4-2x}{3} \\\\ y=\boxed{\sf \frac{2x-4}{3} } \\\\slope = \dfrac{2}{3} \\\\ y-intersept = -\dfrac{4}{3 }\\\\ -----------------$

$\sf \#16 .\\\\ x+y=2x+3 \\\\y=2x-x+3 \\\\\boxed{\sf y=1\cdot x+3} \\\\slope =1 \\\\y-intersept =3$

4 0

2 years ago

6 apple trees fir every 4 pears trees How many apple trees would there be if there were 42 pear trees

SIZIF [17.4K]

Answer: 63 apple trees
Explanation: divide 42 by 4 to get that amount of groups (10.5) then multiply that number by 6 so (10.5 • 6) and that should give the apple tree total.

4 0

3 years ago

What eigen value for this matix <br> (1 -2)<br> (-2 0)

natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

Compute the matrix $A' = A-\lambda I$ , where I is the identity matrix (1s on the diagonal, 0s elsewhere)
Compute the determinant of A'
Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

$A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]$

The determinant of this matrix is

$\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4$

Finally, we have

$\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}$

So, the two eigenvalues are

$\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}$

5 0

3 years ago

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