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3 years ago

One of the solutions to x^2-2x-15=0 is x=-3. What is the other solution?

2 answers:

6 0

X=5 because the factored form of the equation is (x-5)(x+3)=0 so if u set both of these equal to 0, then the other solution is 5.

fiasKO [112]3 years ago

5 0

Answer:

Other solution of the equation is x = 5

Step-by-step explanation:

Given Quadratic Equation,

$x^2-2x-15=0$

One of the solution of the given quadratic equation is x = -3

To find: 2nd solution of the equation.

We solve the given equation using middle term split method.

Consider,

$x^2-2x-15=0$

$x^2-5x+3x-15=0$

x ( x - 5 ) + 3 ( x - 5 ) = 0

( x - 5 )( x + 3 ) = 0

x - 5 = 0 ⇒ x = 5

x + 3 = 0 ⇒ x = -3

Therefore, Other solution of the equation is x = 5

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Need a bit of help with this question:

DiKsa [7]

Firstly let's find hypotenuse(let it will be "n" of smaller triangle

Let use Pythagorean theorem

$a^{2} + {b}^{2} = {n}^{2} \\ {20}^{2} + {10}^{2} = {n}^{2} \\ n = \sqrt{500}$

Now we need to find hypotenuse(x) of bigger triangle

${c}^{2} + {n}^{2} = {x}^{2} \\ {9}^{2} + { \sqrt{500} }^{2} = {x}^{2} \\ x = \sqrt{581}$

The value of x must be rounded to 1 DP, so

$\sqrt{581} = 24.1039... \\ \sqrt{581} \simeq \: 24.1$

Answer: x=24.1

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3 years ago

What is the triangles' perimeter

DENIUS [597]

To find the perimeter of any shape, you find the sum of their side lengths.

In this case, this triangle has side lengths of 1.197, 0.764, and 1.74.
Add these, and the sum is your answer.

1.197 + 0.764 + 1.74 = 3.701

Therefore 3.701 is the correct answer :)

6 0

3 years ago

Read 2 more answers

20 points time cause i appreciate the help

Crazy boy [7]

Answer: 4188.8

Step-by-step explanation:

5 0

3 years ago

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#4 home do I find which one it is?

lina2011 [118]

$\bf \begin{array}{lllll}
solutions&graphs&slopes\\
----&----&----\\
\textit{exactly one}&
\begin{array}{llll}
\textit{the two lines intersect}\\
\textit{at one point}
\end{array}&\textit{different slopes}\\\\
infinitely\quad many&
\begin{array}{llll}
\textit{the two lines coincide}\\
\textit{one is right on top}\\
\textit{ of the other}
\end{array}&
\begin{array}{llll}
\textit{equal slopes}\\
\textit{equal y-intercepts}
\end{array}
\end{array}$

$\bf \textit{no solution}\qquad\quad &\textit{lines are parallel} \qquad &
\begin{array}{llll}
\textit{equal slopes}\\
\textit{different y-intercepts}
\end{array}
\end{array}$

for example, let's look at the first set

y+3x =5 or y = -3x+ 5
and y = -3x + 2
y = m + b

the slopes are equal, the y-intercepts differ
that means, they're just parallel lines, no solution

8 0

3 years ago

susan is buying supplies for a party. if spoons only come in bags of 8 and forks only come in bags of 6, what is the least numbe

Free_Kalibri [48]

Well your are going to find the factors of both 6 and 8 and you will eventually find the the least common multiple for both of them is 24.

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