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ICE Princess25 [194]
3 years ago
13

Please help im stuck on these questions

Mathematics
2 answers:
jeka57 [31]3 years ago
7 0

Answer:

may I help u....................

....... m

Lady bird [3.3K]3 years ago
6 0

Step-by-step explanation:

For no. 3

x - 3 = 3 - x

x + x = 3 + 3

2x = 6

x = 6 / 2

x = 3

For no. 4

3(x + 4) = 3x + 4

3x + 12 = 3x + 4

3x - 3x = 4 - 12

0 = - 8

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I really need help quick!!:) please!! I will give brainliest to first right answer!!!
kykrilka [37]

Answer:

SA = <em>41.</em>

Step-by-step explanation:

Add up everything and you get <em>41 </em>as the SA.

8 0
3 years ago
Someone help me please
Gemiola [76]

Answer:

x=2\sqrt{6}

Step-by-step explanation:

Use the Pythagorean theorem:

a^2+b^2=c^2

a and b are the legs and c is the hypotenuse. Insert the values:

\sqrt{10}^2+\sqrt{14}^2=x^2

Simplify exponents using the rule \sqrt{x}^2=x :

10+14=x^2

Simplify addition:

24=x^2\\x^2=24

Find the square root:

\sqrt{x^2}=\sqrt{24}\\  x=\sqrt{24}

Simplify in radical form: Find a common factor of 24 that is a perfect square:

\sqrt{24}=\sqrt{4*6}

Separate:

\sqrt{4}*\sqrt{6}

Simplify:

2*\sqrt{6}\\x=2\sqrt{6}

Finito.

3 0
3 years ago
Read 2 more answers
the function intersects its midline at (-pi,-8) and has a maximum point at (pi/4,-1.5) write an equation
Tcecarenko [31]

The equation that represents the <em>sinusoidal</em> function is x(t) = -8 + 6.5 \cdot \sin \left[\left(\frac{2}{3} \pm \frac{4\cdot i}{3}\right)\cdot t + \left(\frac{2\pi}{3} \pm \frac{7\pi \cdot i}{3}  \right)\right], i\in \mathbb{Z}.

<h3>Procedure - Determination of an appropriate function based on given information</h3>

In this question we must find an appropriate model for a <em>periodic</em> function based on the information from statement. <em>Sinusoidal</em> functions are the most typical functions which intersects a midline (x_{mid}) and has both a maximum (x_{max}) and a minimum (x_{min}).

Sinusoidal functions have in most cases the following form:

x(t) = x_{mid} + \left(\frac{x_{max}-x_{min}}{2} \right)\cdot \sin (\omega \cdot t + \phi) (1)

Where:

  • \omega - Angular frequency
  • \phi - Angular phase, in radians.

If we know that x_{min} = -14.5, x_{mid} = -8, x_{max} = -1.5, (t, x) = (-\pi, -8) and (t, x) = \left(\frac{\pi}{4}, -1.5 \right), then the sinusoidal function is:

-8 +6.5\cdot \sin (-\pi\cdot \omega + \phi) = -8 (2)

-8+6.5\cdot \sin\left(\frac{\pi}{4}\cdot \omega + \phi \right) = -1.5 (3)

The resulting system is:

\sin (-\pi\cdot \omega + \phi) = 0 (2b)

\sin \left(\frac{\pi}{4}\cdot \omega + \phi \right) = 1 (3b)

By applying <em>inverse trigonometric </em>functions we have that:

-\pi\cdot \omega + \phi = 0 \pm \pi\cdot i, i \in \mathbb{Z} (2c)

\frac{\pi}{4}\cdot \omega + \phi = \frac{\pi}{2} + 2\pi\cdot i, i \in \mathbb{Z} (3c)

And we proceed to solve this system:

\pm \pi\cdot i + \pi\cdot \omega = \frac{\pi}{2} \pm 2\pi\cdot i -\frac{\pi}{4}\cdot \omega

\frac{3\pi}{4}\cdot \omega = \frac{\pi}{2}\pm \pi\cdot i

\omega = \frac{2}{3} \pm \frac{4\cdot i}{3}, i\in \mathbb{Z} \blacksquare

By (2c):

-\pi\cdot \left(\frac{2}{3} \pm \frac{4\cdot i}{3}\right) + \phi =\pm \pi\cdot i

-\frac{2\pi}{3} \mp \frac{4\pi\cdot i}{3} + \phi = \pm \pi\cdot i

\phi = \frac{2\pi}{3} \pm \frac{7\pi\cdot i}{3}, i\in \mathbb{Z} \blacksquare

The equation that represents the <em>sinusoidal</em> function is x(t) = -8 + 6.5 \cdot \sin \left[\left(\frac{2}{3} \pm \frac{4\cdot i}{3}\right)\cdot t + \left(\frac{2\pi}{3} \pm \frac{7\pi \cdot i}{3}  \right)\right], i\in \mathbb{Z}. \blacksquare

To learn more on functions, we kindly invite to check this verified question: brainly.com/question/5245372

5 0
2 years ago
Simplify: | 12^x-8 | = 2sqrt3
lidiya [134]

Answer:

12^x

Step-by-step explanation:

Absolute. Value. Of. |12^x-8|. Is. 12^x + 8

12^x + 8= 2^3.  So. 12^x. +8.=8

12^x+8-8=8-8

12^x=0. Left. With. 12^x?

8 0
3 years ago
What is the axis of symmetry in a parabola khanacademy?
elena-s [515]
The axis of symmetry in a parabola is a line that's vertical that divides into 2 congruent halves. 
7 0
4 years ago
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