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ra1l [238]
3 years ago
11

if your allowance was a penny doubled everyday how many days would it take for you to have over $1000.00

Mathematics
2 answers:
klemol [59]3 years ago
6 0
11days would give you $1024


Day 1:$1
Day2:$2
Day3:$4
Day4:$8
Day5:$16
Day6:$32
Day7:$64
Day8:$128
Day9:$256
Day10:$512
Day11:$1,024
lisabon 2012 [21]3 years ago
4 0

Answer:

it would take 15 days

Step-by-step explanation:

what I did was list them

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Which expression is equivalent to 2^6 x (2^2)^3?<br> O A 212<br> B. 215<br> O c. 236<br> OD 254
Semenov [28]

A. \bold{\boxed{\green{2^{12}}}}

Answer:

Solution given:

2^6 x (2^2)^3

we know that

product of power and power is added and power to power is multiplied for same base only.

I.E.

product of same base but different power:a^b+a^c=a^(b+c)

power to power :(a^b)^c=a^(b*c)

By using above property

2^6+2^{2*3}

2^6+2^6

2^{[6+6]}=2^{12}

8 0
2 years ago
Find the distance between the points S(-5, -2) and T(-3, 4)
Leviafan [203]

Answer:

6.32

Step-by-step explanation:

The x difference:

-5  - -3 = -2

The y difference:

-2 - 4 = -6

Distance:

-6² + -2² = 40

√40 ≅ 6.32

4 0
3 years ago
Which scale of measurement is being used if assigning numbers based on different professional affiliation (e.g., counselors, soc
hammer [34]

Answer: The answer is Nominal scale

Step-by-step explanation:

  • <em>Scale of Measurement</em> refer to ways in which variables/numbers/information are categorized

  • <em>Nominal Scale</em> is a measurement scale, in which numbers serve as “tags”, to identify or classify an object. it is usually use for non-quantitative variables. Some examples of variables that use nominal scales are language spoken, sex, the football club you support, profession etc

8 0
4 years ago
6÷0.9 help plzzzzzzzzzzzzzzzzzzz
Rus_ich [418]

6 / .9 *10/10

60/9

9 goes into 60 6 times with 6 left over

6 6/9

6 2/3

8 0
4 years ago
Read 2 more answers
Could someone please help me:) I am stick and I am not sure what to do ​
Delicious77 [7]

Answer:

Part 5.1.1:

\displaystyle \cos 2A = \frac{7}{8}

Part 5.1.2:

\displaystyle \cos A = \frac{\sqrt{15}}{4}

Step-by-step explanation:

We are given that:

\displaystyle \sin 2A = \frac{\sqrt{15}}{8}

Part 5.1.1

Recall that:

\displaystyle \sin^2 \theta + \cos^2 \theta = 1

Let θ = 2<em>A</em>. Hence:

\displaystyle \sin ^2 2A + \cos ^2 2A = 1

Square the original equation:

\displaystyle \sin^2 2A = \frac{15}{64}

Hence:

\displaystyle \left(\frac{15}{64}\right) + \cos ^2 2A = 1

Subtract:

\displaystyle \cos ^2 2A = \frac{49}{64}

Take the square root of both sides:

\displaystyle \cos 2A = \pm\sqrt{\frac{49}{64}}

Since 0° ≤ 2<em>A</em> ≤ 90°, cos(2<em>A</em>) must be positive. Hence:

\displaystyle \cos 2A = \frac{7}{8}

Part 5.1.2

Recall that:

\displaystyle \begin{aligned}  \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ &=   1- 2\sin^2\theta \\ &= 2\cos^2\theta - 1\end{aligned}

We can use the third form. Substitute:

\displaystyle \left(\frac{7}{8}\right) = 2\cos^2 A - 1

Solve for cosine:

\displaystyle \begin{aligned} \frac{15}{8} &= 2\cos^2 A\\ \\ \cos^2 A &= \frac{15}{16} \\ \\ \cos A& = \pm\sqrt{\frac{15}{16}} \\ \\ \Rightarrow \cos A &= \frac{\sqrt{15}}{4}\end{aligned}

In conclusion:

\displaystyle \cos A = \frac{\sqrt{15}}{4}

(Note that since 0° ≤ 2<em>A</em> ≤ 90°, 0° ≤ <em>A</em> ≤ 45°. Hence, cos(<em>A</em>) must be positive.)

4 0
3 years ago
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