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2 years ago

Help e please i’ll give brainliest

Mathematics

1 answer:

AlladinOne [14]2 years ago

3 0

Answer:

363,000,000

..........

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Which are solutions of x2=-11+4?

Dovator [93]

For this case we have a quadratic equation of the form:

$ax ^ 2 + bx + c = 0\\x ^ 2 + 11x-4 = 0$

Where:

$a = 1\\b = 11\\c = -4$

We find the roots:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}\\x = \frac {-11 \pm \sqrt {11 ^ 2-4 (1) (- 4)}} {2 (1)}\\x = \frac {-11 \pm \sqrt {121 + 16}} {2}\\x = \frac {-11 \pm \sqrt {137}} {2}$

The roots are:

$x_ {1} = \frac {-11+ \sqrt {137}} {2}\\x_ {2} = \frac {-11- \sqrt {137}} {2}$

Answer:

$x_ {1} = \frac {-11+ \sqrt {137}} {2}\\x_ {2} = \frac {-11- \sqrt {137}} {2}$

3 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B45%7D%20%2B%5Csqrt%7B54%7D%20%2B%5Csqrt%7B56%7D%20%3D" id="TexFormula1" title="\sqrt

Artyom0805 [142]

$\sqrt{45}+\sqrt{54}+\sqrt{56}$

$\sqrt{45} = 3\sqrt{5}$

$\sqrt{54} = 3\sqrt{6}$

$\sqrt{56} = 2\sqrt{14}$

<em>so it's equal to:</em>

$=3\sqrt{5}+3\sqrt{6}+2\sqrt{14}$

<h3><em>Hope I helped you!</em></h3><h3><em>Success!</em></h3>

<em>by a random romanian guy</em>

5 0

2 years ago

Read 2 more answers

A total of 6 cups of flour were used to make 4 pans of brownies. Which graph has a

mihalych1998 [28]

I think its option G (if correct plz mark brainliest)

7 0

2 years ago

Determined if the graph represents a polynomial function if it does determine the number of turning points and the least degree

skelet666 [1.2K]

<em>The graph of f is a polynomial function</em>
<em>There are two turning points, namely, x = 0 and x = -2</em>

<h2>Explanation:</h2>

A polynomial function in one variable is given by the form:

$f(x)=a_{n}x^n+a_{n-1}x^{n-1}+\cdots +a_{2}x^2+a_{1}x+a_{0}$

Since you haven't provided any expression, I'll choose the following function:

$f(x)=x^{3}+3x^{2}+3$

So this is indeed a polynomial function. A turning point is an x-value where we have either a local maximum or local minimum. So we need to take the derivative of this functions:

$f'(x)=3x^2+6x \\ \\ Finding \ turning \ points: \\ \\ 3x^2+6x=0 \\ \\ x(3x+6)=0 \\ \\ \\ So: \\ \\ x=0 \\ \\ and \\ \\ \ 3x+6=0 \therefore \ \boxed{x=-2}$