Theses are just some problems i did not understand how to do so if you could please explain and write the answer(:
anything helps
thank you(:
2. Simplify
---(3x^11z^5)^3
b. (3x^5)^ -3 / 2x-5(3y)^4
c.3x^5)^ -3 / 2x ^ -5(3y)^4 (same as b but - 5 is the exponent)
3a.x^2 / 5 + x-5 / 6=0 (proportion, and then i got to the quadratic equation and after i did that i got 5/3 and -2.5 correct?)
3b.18=2(3)^x+3 (x+3 are exponents)
c. 128=4^3x (whats x)
2.
(3x^11z^5)^3 = 27 x^33 z^15
b) and c) are not clear since you don't use proper brackets
3a) again, your lack of brackets make it ambiguous
I will assume you mean
x^2/5 + (x-5)/6 = 0
then
x^2/5 = (5-x)/6
cross-multiply
6x^2 = 25 - 5x
6x^2 + 5x - 25 = 0
your answer is correct
3b)
18 = 2(3)^(x+3)
9 = 3^(x+3)
3^2 = 3^(x+2)
so x+2 = 2
x = 0
c)
128 = 4^(3x)
2^7 = (2^2)^(3x)
2^7 = 2^(6x)
6x = 7
x = 7/6
this is how 2b and 2c is written:
(3x^5)^-3
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2x^-5(3y)^4
Does that help?
(3x^5)^-3
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2x-5(3y)^4
that is what think alyssa is trying to say?
Sure! Let's go through each problem step by step and explain how to solve them:
2. Simplify:
a. To simplify the expression (3x^11z^5)^3, we need to apply the exponent outside the parentheses to every term inside. In this case, we raise each term to the power of 3:
(3x^11z^5)^3 = 3^3 * (x^11)^3 * (z^5)^3 = 27x^33z^15
b. In this problem, we have (3x^5)^-3 divided by 2x^-5(3y)^4. To simplify, we can apply the exponent to each factor:
(3x^5)^-3 / 2x^-5(3y)^4 = (1 / (3x^5)^3) / (2 / x^5)(3y^4) = (1 / 27x^15) / (6x^-5y^4) = 1 / (27x^15 * 6x^-5y^4) = 1 / (162x^10y^4)
c. This problem is similar to part b, but with -5 as the exponent. So we have (3x^5)^-3 / 2x^-5(3y)^4. Using the same steps as before, we get:
(3x^5)^-3 / 2x^-5(3y)^4 = (1 / (3x^5)^3) / (2 / x^-5)(3y^4) = (1 / 27x^15) / (2x^-5 * 3y^4) = (1 / (27x^15 * 2x^-5 * 3y^4) = 1 / (162x^10y^4)
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3a. To solve the equation, x^2 / 5 + x^-5 / 6 = 0, we need to find the values of x that satisfy the equation.
First, we can simplify the fractions by finding a common denominator:
x^2 / 5 + x^-5 / 6 = (6x^2 + 5x^-5) / 30
Now, we solve for x by setting the numerator of the fraction equal to zero:
6x^2 + 5x^-5 = 0
Multiply through by x^5 to get rid of the negative exponent:
6x^7 + 5 = 0
Now we can subtract 5 from both sides:
6x^7 = -5
Divide by 6:
x^7 = -5/6
To solve for x, take the 7th root of both sides:
x = (-5/6)^(1/7)
So, the two possible solutions for x are approximately 0.867 and -0.995.
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3b. The equation is given as 18 = 2(3)^x+3.
To solve for x, we need to isolate it on one side of the equation.
First, divide both sides by 2:
9 = 3(3)^x+3
Next, divide both sides by 3:
3 = (3)^x+3
Now, we can rewrite the equation as an exponent:
3^1 = 3^x+3
Since the bases are equal, the exponents must also be equal. Therefore, we can set x+3 equal to 1:
x+3 = 1
Subtract 3 from both sides:
x = -2
So, the solution is x = -2.
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3c. The equation is given as 128 = 4^(3x).
To solve for x, we need to rewrite the equation using the same base on both sides.
We can rewrite 128 as 2^7:
2^7 = 4^(3x)
Now, we can rewrite 4 as 2^2:
2^7 = (2^2)^(3x)
Using the property of exponents, we can simplify the right side as:
2^7 = 2^(2 * 3x)
Since the bases are equal, the exponents must also be equal. Therefore, we can set 7 equal to 2 * 3x:
7 = 6x
Divide both sides by 6:
x = 7/6
So, the solution is x = 7/6, or approximately 1.1667.