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Agata [3.3K]
3 years ago
13

PLEASE HELP !!!!!!!!!!

Mathematics
1 answer:
Volgvan3 years ago
3 0

Answer:

C.13

Step-by-step explanation:

D is a midpoint, which means that AD=DB= 1/2 AB = 1/2 (24) = 12

Using Pythagorean theorem we know that CD^2+DB^2=CB^2

That means that 5^2+12^2=CB^2

      25+144=169

169=13^2

CB=13

a=CB=13

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What is the final amount if 660 is increased by 1% followed by a further 1% increase? Give your answer rounded to 2 DP.
Delvig [45]

Answer:

$673.27

Step-by-step explanation:

Initial amount = $660

1% Increase = $660*(1+1%)

1% Increase = $660+$6.6

1% Increase = $666.6

Further 1% Increase = $666.6*(1+1%)

Further 1% Increase = $666.6 + $6.666

Further 1% Increase = $673.266

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2 years ago
There are 8 pencils in a package. How many packages will be needed for 28 children if each child gets 4
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Roughly 102 cases is the answear
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3 years ago
Evaluate by using the order of operations: 2+3(5-3)^2÷12
swat32
PEMDAS

2+3(2)^2/12

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3 years ago
What is the length of the the midsegment of a trapezoid with bases of length and ​?
Marina86 [1]

Answer:

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Step-by-step explanation:

3 0
2 years ago
In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini
svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2

The augmented matrix is

\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

R_2\rightarrow R_2-R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]

R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

R_3\rightarrow R_3-R_2

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

k^2-k-2=0

\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2

Case k = −1:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]

This gives the infinite many solution.

5 0
3 years ago
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