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tiny-mole [99]
3 years ago
9

Can someone please help me i have no one to help me

Mathematics
2 answers:
GREYUIT [131]3 years ago
6 0
It is an acute triangle because none of its angles are equal to (right) or above (obtuse) 90 degrees.
Naya [18.7K]3 years ago
3 0
B. Acute triangle. Because all three angles are less than 90 degrees.
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Which option summarizes what gave Chandra Gupta I power
mars1129 [50]

Answer:

Chandra Gupta I, King of India reigned from 320 to about 330 CE and founder of the Gupta empire. He was the grandson of Sri Gupta, the first known leader of the Gupta line. Chandra Gupta I, whose first life is unknown, became the local chief of the kingdom of Magadha parts of the modern state of Bihar.

Step-by-step explanation:

6 0
3 years ago
Katie solved the equation by completing the square. X2+12x−7=0 which equation shows one of the steps katie could have taken to c
sdas [7]

solution:

x^{2}+12x-7=0\\add 7 to both side of the equation\\x^{2}-12x=7\\to creat a trinomial square on the left side of the equation.\\find a value that is equal to the square of b and the cofficent of x.\\(\frac{b}{2})^2=(-6)^2\\add the term to each side of the equation\\x^2-12x+(-6)^2=7+(-6)\\x^2-12x+36=43\\(x-6)^2=43\\solve the equation for x.\\take the square root of each side of the equation to the set up the solution x,\\(x-6)^2.\frac{1}{2}=\pm \sqrt{43}\\x-6=\pm \sqrt{43}\\x=6\pm \sqrt{43}

4 0
3 years ago
Read 2 more answers
If (x^2−1)/(x+1) = 3x + 5, then x + 3 =<br> (A) -3<br> (B) -2<br> (C) 0<br> (D) 2<br> (E) 4
sineoko [7]
The answer is C) 0
Because math
4 0
3 years ago
6 1/2 - 7/8 : 5 11/16
Komok [63]

Answer:

Step-by-step explanation:

brainly.com/question/6752935

hope this helps

7 0
3 years ago
3x^2+kx=-3 What is the value of K will result in exactly one solution to the equation?
Scorpion4ik [409]

Answer:

For k = 6 or k = -6, the equation will have exactly one solution.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = (x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

If \bigtriangleup = 0, the equation has only one solution.

In this problem, we have that:

3x^{2} + kx + 3 = 0

So

a = 3, b = k, c = 3

\bigtriangleup = b^{2} - 4ac

\bigtriangleup = k^{2} - 4*3*3

\bigtriangleup = k^{2} - 36

We will only have one solution if \bigtriangleup = 0. So

\bigtriangleup = 0

k^{2} - 36 = 0

k^{2} = 36

k = \pm \sqrt{36}

k = \pm 6

For k = 6 or k = -6, the equation will have exactly one solution.

3 0
3 years ago
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