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3 years ago

I need the answers for 1 a-d pleas

Mathematics

1 answer:

DaniilM [7]3 years ago

4 0

We want to find a fraction that is between

$\frac{1}{4}\:and\:\frac{1}{2}$

There are several ways to go about this.

There are also infinitely many fractions that can be found between the given fractions.

Let us go for the one in the middle of the two fractions.

So we add the two fractions and divide by 2.

a) $\frac{\frac{1}{4} +\frac{1}{2} }{2}$

Find LCM for the numerator

$\frac{\frac{1+2}{4}}{2}$

Simplify:

$\frac{\frac{3}{4}}{2}$

$\frac{\frac{3}{2}}{2}$

$\frac{3}{8}}$

b) $\frac{\frac{1}{3} +\frac{3}{4} }{2}=\frac{13}{24}$

c) $\frac{\frac{1}{2} +\frac{3}{4} }{2}=\frac{5}{8}$

d) $\frac{\frac{3}{5} +\frac{7}{8} }{2}=\frac{59}{80}$

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State the system of linear inequalities which satisfies the following shaded regions. help ??asappp

7nadin3 [17]

Find the equation of the line that does not have an equation written for it.

(4, -4) (0,-2)

m = (-2 + 4) / (0 - 4)

m = 2 / -4 = -1/2

y = -1/2x - 2

Let's now look at all of these different lines we have. For the two intersecting lines, we know that the bottom one must be greater than and the top must be less than. And because these are solid lines, they are greater/less than or equal to. For the dotted line, we know that x must be less than because the line is dotted and the shading is to the left.

y ≤ x - 2

x < 4

y ≥ -1/2x - 2

Hope this helps! :)

6 0

3 years ago

Find c1 and c2 such that M2+c1M+c2I2=0, where I2 is the identity 2×2 matrix and 0 is the zero matrix of appropriate dimension.

Katyanochek1 [597]

The question is missing parts. Here is the complete question.

Let M = $\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]$ . Find $c_{1}$ and $c_{2}$ such that $M^{2}+c_{1}M+c_{2}I_{2}=0$ , where $I_{2}$ is the identity 2x2 matrix and 0 is the zero matrix of appropriate dimension.

Answer: $c_{1} = \frac{-16}{10}$

$c_{2}=\frac{-214}{10}$

Step-by-step explanation: Identity matrix is a sqaure matrix that has 1's along the main diagonal and 0 everywhere else. So, a 2x2 identity matrix is:

$\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

$M^{2} = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]$

$M^{2}=\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]$

Solving equation:

$\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+c_{1}\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right] +c_{2}\left[\begin{array}{cc}1&0\\0&1\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

Multiplying a matrix and a scalar results in all the terms of the matrix multiplied by the scalar. You can only add matrices of the same dimensions.

So, the equation is:

$\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$