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3 years ago

These triangles are similar. True Or False

Mathematics

1 answer:

iren2701 [21]3 years ago

3 0

Answer:

True because they both are triangles, they both have tree side but one is just smaller than the other.

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What value of x will make the equation below true? 14−(x+7)=3

zalisa [80]

14-x-7=3
7-x=3
-x=-4
x=4

7 0

2 years ago

Write -4 3/8 as a decimal

Likurg_2 [28]

-4 x 8= -32+3=-29. -29/8= -3.625

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2 years ago

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Divide using long division. (9p2 + 8p3 + 12) ÷ (p + 1)

kodGreya [7K]

Answer:

8p∧2 ± p - 1 ± 13/p ± 1

Step-by-step explanation:

Divide (9p2 + 8p3 + 12) ÷ (p + 1) and it equals 8p∧2 ± p - 1 ± 13/p ± 1

3 0

2 years ago

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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

2 years ago

Factor completely and then place the factors in the proper location on the grid. 36x 2 + 60x + 25

Aleonysh [2.5K]

Since 36x² + 60x + 25 is the perfect square trinomial, we use this formula a² + 2ab + b² = (a + b)². Hence, the solution is (6x + 5)²

4 0

2 years ago

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