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elena-s [515]
3 years ago
7

he closing price of a share of stock in Company XYZ is $25.69 on Thursday. If the change from the closing price on Wednesday is

–$0.75, find the closing price on Wednesday. $24.94 $25.75 $26.44 $25.06
Mathematics
1 answer:
mamaluj [8]3 years ago
6 0

Answer:

Option C - $26.44

Step-by-step explanation:

To find the closing proce on Wednesday

Company XYZ is $ 25.69 on Thursday

Company XYZ is - $0.75 on Wednesday

This means the calculation should be as following,

25.69 + 0.75 = 26.44$ the closing price for Wednesday

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Describe how solving 1x + 4 = 0 is similar to solving<br> 1x + 4 = 0.
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They are both the same equations, the procedure to obtain x value will be equivalent.

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Sam's Office Supply store has 436 pens.Sam orders 6 pens each in 8 different colors to add to his store. How many pens did Sam o
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48 pens

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sam ordered 48 pens

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2 years ago
Which could be the first step in simplifying this expression? Check all that apply. (x cubed x Superscript negative 6 Baseline)
PtichkaEL [24]

Answer:

(x^{-3} )^{2}

x^6 x^{-12}

Step-by-step explanation:

(x^{3} x^{-6} )^{2} is the expression given to be solved.

First of all let us have a look at <u>3 formulas</u>:

1.\ p^a \times p^b = p^{(a+b)}\\2.\ (p^a \times q^b)^c = (p^{a})^c \times (q^{b})^c\\3.\ (p^a)^b = p^{a\times b}

Both the formula can be applied to the expression((x^{3} x^{-6} )^{2}) during the first step while solving it.

<u>Applying formula (1):</u>

(x^{3} x^{-6} )^{2}

Comparing the terms of (x^{3} x^{-6} ) with p^a \times p^b

p=x, a =3, b=-6

\Rightarrow x^{3+(-6)}\\\Rightarrow x^{3-6}\\\Rightarrow x^{-3}

So, (x^{3} x^{-6} )^{2} is reduced to (x^{-3} )^{2}

<u>Applying formula (2):</u>

Comparing the terms of (x^{3} x^{-6} )^{2} with (p^a \times q^b)^c

p=q=x, a =3, b=-6, c=2

\Rightarrow (x^{3})^2\times (x^{-6})^2\\\text{Applying Formula (3)}\\x^6 x^{-12}

So, (x^{3} x^{-6} )^{2} is reduced to x^6 x^{-12}.

So, the answers can be:

(x^{-3} )^{2}

x^6 x^{-12}

8 0
3 years ago
A sample of radium has a weight of 1.5 mg and a half-life of approximately 6 years.
pogonyaev

Answer:

0.75 mg

Step-by-step explanation:

From the question given above the following data were obtained:

Original amount (N₀) = 1.5 mg

Half-life (t₁/₂) = 6 years

Time (t) = 6 years

Amount remaining (N) =.?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t₁/₂) = 6 years

Time (t) = 6 years

Number of half-lives (n) =?

n = t / t₁/₂

n = 6/6

n = 1

Finally, we shall determine the amount of the sample remaining after 6 years (i.e 1 half-life) as follow:

Original amount (N₀) = 1.5 mg

Half-life (t₁/₂) = 6 years

Number of half-lives (n) = 1

Amount remaining (N) =.?

N = 1/2ⁿ × N₀

N = 1/2¹ × 1.5

N = 1/2 × 1.5

N = 0.5 × 1.5

N = 0.75 mg

Thus, 0.75 mg of the sample is remaining.

5 0
2 years ago
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