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sergejj [24]
2 years ago
12

For what value of y does 125 = (StartFraction 1 Over 25 EndFraction) Superscript y minus 1? Negative five-halves –2 –1 Negative

one-half.
Mathematics
1 answer:
rodikova [14]2 years ago
7 0

Answer:

D. -1/2

Step-by-step explanation:

First to rewrite the equation in numbers

125=\frac{1}{25}^{y-1}

First, we need to find what exponent would get 1/25 to 125.

\frac{1}{25}^{y}=125

Sense 1/25 is a fraction we must use a negative exponent to make it a whole number. We can just find what exponent would get 25 to 125 then make it negative. 1 wouldn't be enough and 2 would be too much so y is between 1 and 2 into negative is between -1 and -2. Sense the original equation has (y-1) the only option you can subtract 1 from to get between -1 and -2 is -1/2 which would give -1 1/2.

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Can some please help me with Algebra 2 work? Its problems 45-48 all and 53-57 just odd numbers
Elanso [62]

45)

E = mc²     <em> (solve for "m" by dividing both sides by c²)</em>

E/c² = m

46)

c(a + b) - d = f  <em>(to solve for "a"; add "d", divide by "c" and then subtract "b")</em>

c(a + b)     = f + d

(a + b)       = (f + d)/c

a              = (f + d)/c  - b

47)

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z/πq³ = h

48)

(x + y)/z - a = b   <em> (solve for "y"; add "a", multiply by "z", and then subtract "x")</em>

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(x + y)         = z(b + a)

     y           = z(b + a) - x

53)

5x - 9 = 11x + 3

     -9 = 6x + 3      <em>subtracted 5x from both sides</em>

    -12 = 6x          <em>subtracted 3 from both sides</em>

      -2 = x           <em>divided both sides by 6</em>

55)

5.4(3k - 12)   +   3.2(2k + 6) = -136

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22.6k - 45.6 = -136             <em>added like terms (16.2k + 6.4k and -64.8 + 19.2</em>

22.6k = -90.4                    <em>added 45.6 to both sides</em>

      k = -4                       <em>divided both sides by 22.6</em>

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(4/9)y + 5 = (-7/9)y - 8

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8 0
4 years ago
3log54+5log50-3log2-5log2
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4 0
4 years ago
The height h of a projectile is a function of the time t it is in the air. the height in feet for t seconds is given by the func
alexgriva [62]

Domain means the values of independent variable(input) which will give defined output to the function.

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The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function

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To get defined output, the height h(t) need to be greater than or equal to zero. We need to set up an inequality and solve it to find the domain values.

To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq  0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq  0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible  \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)

Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq  0\\-112 \geq  0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq  0\\80 \geq  0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq  0\\-112 \geq  0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution

Conclusion:

The domain of the function is the time in between 0 to 6 seconds

0 \leq  t \leq  6

The height will be positive in the above interval.

7 0
3 years ago
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