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Hitman42 [59]
3 years ago
5

Which statement is true for the equation 5n − 4 = 5n − 3?

Mathematics
1 answer:
Molodets [167]3 years ago
8 0

Answer:

no solution

Step-by-step explanation:

  1. add 4 to each side
  2. subtract 5n to both sides
  3. answer 0=1
  4. no solution

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Evaluate each logarithm. Do not use a calculator. Log (125)(1/25)
svetlana [45]

Answer:

-2/3

Step-by-step explanation:

Log125(1/25)=x

Raise each side to the base of 125

125^Log125(1/25)=125^x

1/25 = 125^x

Rewrite 25 as a power of 5  and 125 as a power of 5

1 / 5^2 = 5^3^x

The if power is in the denominator, we can bring it to the numerator by making it negative

5^-2 = 5^3^x

We know that a^b^c = a^(b*c)

5^-2 = 5^(3*x)

Since the bases are the same, the exponents are the same

-2 = 3x

Divide by 3

-2/3 = 3x/3

-2/3 =x

6 0
3 years ago
Please help if possible its important!!​
Otrada [13]

Answer:

600 ft³

Step-by-step explanation:

Base area: B

Base is triangle

base = 12 ft and height = 5 ft

Area of triangle = B = \frac{1}{2}* \ base * \  height

                                 = \frac{1}{2}*12 * 5\\\\= 6 * 5\\\\= 30 \ ft^{2}

H = 20 ft

Volume = B * H

             = 30 * 20

             = 600 ft³

7 0
3 years ago
Which statements are true about the graph of the function f(x) = x2 – 8x + 5? Check all that apply.
statuscvo [17]

Answer:

A, D, E are true

Step-by-step explanation:

You have to complete the square to prove A.  Do this by first setting the function equal to 0, then moving the 5 to the other side.

x^2-8x=-5

Now we can complete the square.  Take half the linear term, square it, and add it to both sides.  Our linear term is 8 (from the -8x).  Half of 8 is 4, and 4 squared is 16.  So we add 16 to both sides.

(x^2-8x+16)=-5+16

We will do the addition on the right, no big deal.  On the left, however, what we have done in the process of completing the square is to create a perfect square binomial, which gives us the h coordinate of the vertex.  We will rewrite with that perfect square on the left and the addition done on the right,

(x-4)^2=11

Now we will move the 11 back over, which gives us the k coordinate of the vertex.

(x-4)^2-11=y

From this you can see that A is correct.

Also we can see that the vertex of this parabola is (4, -11), which is why B is NOT correct.

The axis of symmetry is also found in the h value.  This is, by definition, a positive x-squared parabola (opens upwards), so its axis of symmetry will be an "x = " equation.  In the case of this type of parabola, that "x = " will always be equal to the h value.  So the axis of symmetry is

x = 4, which is why C is NOT correct, either.

We can find the y-intercept of the function by going back to the standard form of the parabola (NOT the vertex form we found by completing the square) and sub in a 0 for x.  When we do that, and then solve for y, we find that when x = 0, y = 5.  So the y-intercept is (0, 5).

From this you can see that D is also correct.

To determine if the parabola has real solutions (meaning it will go through the x-axis twice), you can plug it into the quadratic formula to find these values of x.  I just plugged the formula into my graphing calculator and graphed it to see that it did, indeed, go through the x-axis twice.  Just so you know, the values of x where the function go through are (.6833752, 0) and (7.3166248, 0).  That's why you need the quadratic formula to find these values.

7 0
3 years ago
Alexis is making goodie bags for her birthday party. She has made 12 bags already.
Genrish500 [490]
20
Needs 20 characters
8 0
3 years ago
Your grandparents invested $2,000 for you on the day you were born. How much will this investment be worth on your 25th birthday
algol [13]

We know that, Amount in Compound interest is given by :

\bigstar \ \ \boxed{\sf{Amount = Principal\bigg(1 + \dfrac{Rate \ of \ Interest}{100}\bigg)^{Time \ Period}}}

Given : Principal = $2000

Given : Annual yield is 5% and the interest is compounded quarterly

It means : Interest is compounded 4 times in a year

\implies \sf{Rate \ of \ Interest = \dfrac{R}{4} = \dfrac{5}{4}}

\sf{\implies Time \ period = (25 \times 4) = 100}

Substituting all the values in the formula, we get :

\implies \sf{Amount = 2000\bigg(1 + \dfrac{\dfrac{5}{4}}{100}\bigg)^{100}}

\implies \sf{Amount = 2000\bigg(1 + \dfrac{5}{400}\bigg)^{100}}

\implies \sf{Amount = 2000\bigg(1 + \dfrac{1}{80}\bigg)^{100}}

\implies \sf{Amount = 2000\bigg(\dfrac{81}{80}\bigg)^{100}}

\implies \sf{Amount = 2000 \times (1.0125)^{100}}

\implies \sf{Amount = 2000 \times 3.463}}

\implies \sf{Amount = 6926.8}

8 0
3 years ago
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