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Zielflug [23.3K]
3 years ago
15

Before school, janine spends 1/10 hour making the bed, 1/5 hour getting dressed, and 1/2 hour eating breakfast. What fraction of

an hour does she spend doing these activities
Mathematics
1 answer:
N76 [4]3 years ago
6 0
4/5 of an hour spent
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Find the Area Please​
GuDViN [60]
12x8=96
This should be the correct answer
6 0
2 years ago
Determine whether each expression is equivalent to 49^2t – 0.5.
vampirchik [111]

Answer:

None of the expression are equivalent to 49^{(2t - 0.5)}

Step-by-step explanation:

Given

49^{(2t - 0.5)}

Required

Find its equivalents

We start by expanding the given expression

49^{(2t - 0.5)}

Expand 49

(7^2)^{(2t - 0.5)}

7^2^{(2t - 0.5)}

Using laws of indices: (a^m)^n = a^{mn}

7^{(2*2t - 2*0.5)}

7^{(4t - 1)}

This implies that; each of the following options A,B and C must be equivalent to 49^{(2t - 0.5)} or alternatively, 7^{(4t - 1)}

A. \frac{7^{2t}}{49^{0.5}}

Using law of indices which states;

a^{mn} = (a^m)^n

Applying this law to the numerator; we have

\frac{(7^{2})^{t}}{49^{0.5}}

Expand expression in bracket

\frac{(7 * 7)^{t}}{49^{0.5}}

\frac{49^{t}}{49^{0.5}}

Also; Using law of indices which states;

\frac{a^{m}}{a^n} = a^{m-n}

\frac{49^{t}}{49^{0.5}} becomes

49^{t-0.5}}

This is not equivalent to 49^{(2t - 0.5)}

B. \frac{49^{2t}}{7^{0.5}}

Expand numerator

\frac{(7*7)^{2t}}{7^{0.5}}

\frac{(7^2)^{2t}}{7^{0.5}}

Using law of indices which states;

(a^m)^n = a^{mn}

Applying this law to the numerator; we have

\frac{7^{2*2t}}{7^{0.5}}

\frac{7^{4t}}{7^{0.5}}

Also; Using law of indices which states;

\frac{a^{m}}{a^n} = a^{m-n}

\frac{7^{4t}}{7^{0.5}} = 7^{4t - 0.5}

This is also not equivalent to 49^{(2t - 0.5)}

C. 7^{2t}\ *\ 49^{0.5}

7^{2t}\ *\ (7^2)^{0.5}

7^{2t}\ *\ 7^{2*0.5}

7^{2t}\ *\ 7^{1}

Using law of indices which states;

a^m*a^n = a^{m+n}

7^{2t+ 1}

This is also not equivalent to 49^{(2t - 0.5)}

6 0
3 years ago
HELP with these questions
zlopas [31]

<u>Step-by-step explanation:</u>

transform the parent graph of f(x) = ln x        into f(x) = - ln (x - 4)  by shifting the parent graph 4 units to the right and reflecting over the x-axis

(???, 0): 0 = - ln (x - 4)

            \frac{0}{-1} = \frac{-ln (x - 4)}{-1}

            0 = ln (x - 4)

            e^{0} = e^{ln (x - 4)}

             1 = x - 4

          <u> +4 </u>  <u>    +4 </u>

             5 = x

(5, 0)

(???, 1): 1 = - ln (x - 4)

            \frac{0}{-1} = \frac{-ln (x - 4)}{-1}

            1 = ln (x - 4)

            e^{1} = e^{ln (x - 4)}

             e = x - 4

          <u> +4 </u>   <u>    +4 </u>

         e + 4 = x

          6.72 = x

(6.72, 1)

Domain: x - 4 > 0

                <u>  +4 </u>  <u>+4  </u>

               x       > 4

(4, ∞)

Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that

No vertical asymptotes

*************************************************************************

transform the parent graph of f(x) = 3ˣ        into f(x) = - 3ˣ⁺⁵  by shifting the parent graph 5 units to the left and reflecting over the x-axis

Domain: there is no restriction on x so domain is all real number

(-∞, ∞)

Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0.  the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.

(-∞, 0)

Y-intercept is when x = 0:

f(x) = - 3ˣ⁺⁵

      = - 3⁰⁺⁵

      = - 3⁵

      = -243

Horizontal Asymptote: y = 0  <em>(explanation above)</em>

5 0
3 years ago
A department store is having a sale in which every item is 38% off its original price. The regular price of a suit is $112. How
lara [203]

Answer: $42.56

Step-by-step explanation:

Percent discount in department store = 38%

The regular price of suit = $112.

Price taken off the regular price of the suit =

38% of $112

So, 38% of $112

= (38/100) x $112

= 0.38 x $112

= $42.56

(i.e the customer will pay $112 - $42.56 = $69.44 only for the suit)

Thus, the price taken off the regular price of the suit is $42.56

3 0
3 years ago
If a point (-3/5,b) is on the graph of the equation 2x+3y=6 and also on the graph of y=x+3, what is the value of b?
AleksAgata [21]
Since the point satisfies the equation
&nbsp; y = x +3
you have
&nbsp; b = (-3/5) +3 = 12/5 . . . . . . . matches A. 12/5
7 0
3 years ago
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