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3 years ago

Solve the equation: log4(x + 20) = 3

Mathematics

2 answers:

-Dominant- [34]3 years ago

8 0

Answer To The Whole Assignment

1. x= 44

2. x=7

3. (in order first box to last) 2,5,1,4,3

4. x= -5

5. D. This number is a true solution of the original equation.

6. x=2

7. A. log2[x(x – 6)] = 4

8. C. x2 – 6x – 16 = 0

9. x=8

10. x=1, x=2

11. There is no solution.

12. x=3 or x=-3

13. C. Only –3 is an extraneous solution.

14. The bases of the logarithms are not the same.The one-to-one property does not apply when the bases are not the same.The change of base formula should have been used to write the logarithms with the same base.

**** all answers were found off of brainly.com *******

Angelina_Jolie [31]3 years ago

6 0

Loga x=n ⇔a^n=x
Therefore:

log₄ (x+20)=3
3⁴=x+20
64=x+20
x=64-20
x=44

Answer: x=44

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In the app you are creating, a duck is going to fly from the coordinates (1, 3) then go 3 units right. And 6 units down. Write a

quester [9]

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y=-2x+5. (4,-3)

Step-by-step explanation:

1) What's being asked in other words is to write a function. So the initial point is (1,3). In x-axis, a translation movement can only be to the right or to the left. Similarly for y-axis the possible movements are shifting up or down. The question says, 3 units right for x-cordinate and 6 units down for y cordinate. So the final point of this translation would be (1+3,3-6) =(4,-3)

2) To write a linear function we need to find out the slope:

$m=\frac{-3-3}{4-1}=m=\frac{-6}{3} m=-2$

$3=1(-2)+b\\3+2=b\\ b=5\\ \\-3=4(-2)+b\\ b=8-3\\ b=5$

y=-2x+5

<u>Proof:</u>

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y=-2(1)+5

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3 years ago

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2 years ago

When a gray kangaroo jumps, its path through the air can be modeled by y= -0.0267x2 + 0.8x where x is the kangaroo's horizontal

VikaD [51]

A gray kangaroo jumped and it's path through the air is given by

y = $-0.0267 x^2 +0.8 x$

where x is the kangaroo's horizontal distance traveled (in feet) wand y is the corresponding height (in feet).

For maximum distance we have to differentiate this expression.

$\frac{\mathrm{d}y }{\mathrm{d} x}=-0.0267\frac{\mathrm{d} x^2}{\mathrm{d} x}+0.8\frac{\mathrm{d} x}{\mathrm{d} x}\\ \frac{\mathrm{d} y}{\mathrm{d} x}=-0.0267\times2x + 0.8$

For maxima or minima

$\frac{\mathrm{d} y}{\mathrm{d} x}$ =0

-0.0267×2 x + 0.8=0

⇒-0.0534 x + 0.8=0

⇒0.8=0.0534 x

⇒x=0.8/0.0534

x =14.98 (approx)

Now differentiating the expression again

$\frac{\mathrm{d^{2}{y}} }{\mathrm{d} x^{2}}$ =-0.0534

Since double derivative is negative , so

x=14.98 will be the point of Maxima.

The Kangaroo can go the maximum distance of 14.98 feet (approx)

and the height that can kangaroo go through=-0.0267×14.98×14.98+0.8×14.98

= -5.9914 +11.984

= 5.9925 feet

= 6 feet (approx)

So, X=Horizontal distance covered=14.98×2=29.96 feet,[∵ at x=14.98 it attains maximum height, So total distance Travelled by kangaroo will be two times of the distance covered that it has gone through to achieve maximum height.]⇒ The path is parabolic. = 30 feet (approx)

Y= Vertical distance covered= 5.9925=6 feet(approx)

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3 years ago