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3 years ago

9

2x − y = −4→10x − 5y = −20 3x 5y = 59→3x 5y = 59 13x = 39 which equation can replace 3x 5y = 59 in the original system and still

produce the same solution? 2x – y = –4 10x – 5y = –20 7x = 39 13x = 39

1 answer:

malfutka [58]3 years ago

8 0

13x = 39 can replace 3x + 5y = 59 in the original system and still produce the same solution.

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What are all the roots of the equation x^2+6x-9

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Answer:

x = - 3 ± 3 $\sqrt{2}$

Step-by-step explanation:

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We can solve for x using the quadratic formula

x = ( - b ± $\sqrt{b^2-4ac }$ ) / 2a

with a = 1, b = 6 and c = - 9

x = ( - 6 ± $\sqrt{6^2-(4(1)(-9)}$ ) / 2

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3 years ago

Find the length of each side of the triangle determined by the three points P1,P2, and P3. State whether the triangle is an isos

Tanya [424]

Answer:

The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$P_{1} = (- 1, 4)$

$P_{2} = (6, 2)$ and

$P_{3} = (4, - 5)$

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+} \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}}$

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Distance between $P_1$ and $P_2$ , (Side 2):

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$(x_2, y_2) = (4, - 5)$

Distance = $\sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2$

$= \sqrt{ 25 + 81 }$

$= \sqrt{\textbf{106}}$

Length of Side 2 = $\sqrt{\textbf{106}}$ units.

Distance between $P_2$ and $P_3$ , Side 3 :

$(x_1, y_1) = (4, 5)$

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}$

Length of Side 3 = $\sqrt{\textbf{53}}$ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

(Side 1)² + (Side 3)² = (Side 2)²

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3 years ago

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Answer:

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Step-by-step explanation:

(x,y) reflected over y axis ---> (-x,y)

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2 years ago

Suppose you can factor x^2 +bx +c as (x+p)(x+q) . If c<0, what could be the possible values of p and q?

Anna11 [10]

Ans: Option A

Explanation:
Let's solve it smartly!
Given expression: $x^{2} + bx +c$ --- (A)
Factors: (x+p)(x+q)
Condition: c<0
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=> $x^{2} + (p+q)x + pq$ --- (B)

By comparing (B) with (A), we can say that:
pq = c --- (C)

Now, as the condition says, c<0, it means either p or q is negative. Both cannot be positive or both cannot be negative.

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Hence, option B and D are wrong.

2) If p<0, q<0 it means c>=0 since (-p)(-q) = (+c)(according to equation (C)). Condition is not met.
Hence option C is out as well.

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Condition is MET!
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3 years ago

Read 2 more answers

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Kazeer [188]

Answer:

Step-by-step explanation:

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3 years ago