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DiKsa [7]
3 years ago
13

How do you solve 25x = (1/125)^(4x-5)

Mathematics
2 answers:
amm18123 years ago
6 0

Step-by-step explanation:

Step 1: 25*x-((1/125)^(4*x-5))=0

step 2: simplify 1/125

step 3: 25x-(1/125)(4x-5)=0

pls i need one more brainly

jeyben [28]3 years ago
6 0

Answer:Step 1: 25*x-((1/125)^(4*x-5))=0

step 2: simplify 1/125

step 3: 25x-(1/125)(4x-5)=0

pls i need two more brainly

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If one movie ticket cost $15, how much will 5 movie tickets cost
Anettt [7]

Answer:

$75

Step-by-step explanation:

15 x 5 = 75

trust me bro :)

7 0
3 years ago
Solve for 50 points<br> x^(2)=100
Eddi Din [679]

x²=100

x²=10×10

x=10

hope it helps you

7 0
3 years ago
Read 2 more answers
In 88 minutes, he uses 132 balloons to make 22 identical balloon sculptures. How many minutes does it take to make one balloon s
Temka [501]

Answer:

a) How many minutes does it take to make one balloon sculpture?

= 4 minutes

b) How many balloons are used in one sculpture?

= 6 balloons

Step-by-step explanation:

In 88 minutes, he uses 132 balloons to make 22 identical balloon sculptures.

a) How many minutes does it take to make one balloon sculpture?

22 identical balloon sculptures = 88 minutes

1 balloon sculpture = x

x = 88 minutes/22

x = 4 minutes

b) How many balloons are used in one sculpture?

22 balloon sculptures = 132 balloons

1 balloon sculpture = x

Cross Multiply =

22x = 132 balloons

x = 132 balloons/22

x = 6 balloons

8 0
3 years ago
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
Pick the correct equation for the line below:
Roman55 [17]

Answer:

none of above because the correct equation is y= -3x+90

6 0
3 years ago
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