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4 years ago

8

11.

1 answer:

Mrrafil [7]4 years ago

8 0

Answer:

2 should be distributed as 2y + 16; y = 8

Step-by-step explanation:

We have to distribute 2, then:

Now we have to sum (-2y) in both sides of the equation:

Finally we have to divide both sides of the equation in 2:

Then the answer is 2 should be distributed as 2y + 16; y = 8

Hope this helps.

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4 years ago

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find two positive even consecutive integers such that a square of the smaller integer is 10 more than the larger integer

nalin [4]

Answer:

the two positive consecutive integers are 4 and 6.

Step-by-step explanation:

Let the smaller integer be s; then s^2 = (s + 2) + 10.

Simplifying, s^2 - s - 2 - 10 = 0, or

s^2 - s - 12 = 0.

Solve this by factoring: (s - 4)(s + 3) = 0.

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The smaller integer is 4. Does this satisfy the equation s^2 = (s + 2) + 10?

4^2 = (4 + 2) + 10 True or False?

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6 0

4 years ago

White raven [17]

Answer:

(a) a=6 and b≠ $\frac{11}{4}$

(b)a≠6

(c) a=6 and b= $\frac{11}{4}$

Step-by-step explanation:

writing equation in agumented matrix form

$\begin{bmatrix}1 &1 & 0 &1 &b\\ 2 &3 & 1 &5 &6\\ 0& 0 & 1 &1 &4\\ 0& 2 & 2&a &1\end{bmatrix}$

now $R_{2} =R_{2}-2\times R_{1}$

$\begin{bmatrix}1 &1& 0 &1 &b\\ 0 &1& 1 &3 &6-2b\\ 0& 0 & 1 &1 &4\\ 0& 2 & 2&a &1\end{bmatrix}$

now $R_{4} =R_{4}-2\times R_{2}$

$\begin{bmatrix}1 &1& 0 &1 &b\\ 0 &1& 1 &3 &6-2b\\ 0& 0 & 1 &1 &4\\ 0& 0 & 0 &a-6 &4b-11\end{bmatrix}$

a) now for inconsistent

rank of augamented matrix ≠ rank of matrix

for that a=6 and b≠ $\frac{11}{4}$

b) for consistent w/ a unique solution

rank of augamented matrix = rank of matrix

a≠6

c) consistent w/ infinitely-many sol'ns

rank of augamented matrix = rank of matrix < no. of variable

for that condition

a=6 and b=[tex]\frac{11}{4}

then rank become 3 which is less than variable which is 4.

4 0

4 years ago