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Reil [10]
3 years ago
11

"( \frac{1}{2})^{2} " align="absmiddle" class="latex-formula">
evaluate
can u show me the steps please
Mathematics
1 answer:
Vladimir79 [104]3 years ago
5 0
All you need to do is solve it
\frac{1}{2} ^{2} or \frac{1}{2}  x \frac{1}{2}
When solved your answer is 0.25 or 1/4
------------------------
What I did to solve it was....

1. Divide top fraction by bottom fraction to get decimal
1 ÷ 2 = 0.5

2. Then multiply 0.5 by its self (the same as putting it to the ^2)
0.5 x 0.5 = 0.25
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Check all of the polynomial functions that have 2 as a root.
Leviafan [203]

A,C,D those were the answers i put and they were correct.

5 0
2 years ago
the length of the sides of a square are initially 0 cm and increase at a constant rate of 11 cm per second. suppose the function
Likurg_2 [28]

The function of the area of the square is A(t)=121t^{2}

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

\frac{dx}{dt} = 11 \frac{cm}{sec}

⇒x=11t

Area of square=(length of side)^{2}

Area of square=(11t)^{2}{as the length of side is 11t}{varies by time}

Area of square=121t^{2}

Therefore,The function of the area of the square is A(t)=121t^{2}

Learn more about The function of the area of the square is A(t)=121t^{2}

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

\frac{dx}{dt} = 11 \frac{cm}{sec}

⇒x=11t

Area of square=(length of side)^{2}

Area of square=(11t)^{2}{as the length of side is 11t}{varies by time}

Area of square=121t^{2}

Therefore,The function of the area of the square is A(t)=121t^{2}

Learn more about area here:

brainly.com/question/27683633

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3 0
1 year ago
Establish the identity.<br> (2 cos 0-6 sin 0)² + (6 cos 0+2 sin 0)2 = 40
kogti [31]

Rewriting the left-hand side as follows,

(2\cos\theta-6\sin \theta)^2 +(6\cos \theta+2\sin \theta)^2\\\\=4\cos^2 \theta-24\cos \theta \sin \theta+36 \sin^2 \theta+36 \cos^2 \theta+24 \cos \theta \sin \theta+4 \sin^2 \theta\\\\=40\cos^2 \theta+40 \sin^2 \theta\\\\=40(\cos^2 \theta+\sin^2 \theta)\\\\=40

8 0
1 year ago
What is 39.79949748 rounded to the nearest hundredth?
Rufina [12.5K]

Answer:

Number = 39.80

Step-by-step explanation:

Given

Number = 39.79949748

Required

Approximate (to the nearest 100th)

This means that, we approximate at the second digit after the decimal.

So:

i.e,

Number = 39.79  [Begin  approximation] 949748

The first digit after [Begin approximation] is then approximated using the following rule:

0 - 4 \approx 0

5 - 9 \approx 1\\

Since 9 falls in 5 - 9 \approx 1\\ category, the number becomes:

Number = 39.[79+1]

Number = 39.80

3 0
2 years ago
Let f(t) give the number of liters of fuel oil burned in t days, and w(r) the liters burned in r weeks. Find a formula for w by
viktelen [127]

Answer:

7 f(t)

Step-by-step explanation:

So, our f(t) is the number of liters burned in t days. If t is 1, f(t)=f(1) and so on for every t.

w(r) id the number of liters in r weeks. This is, in one week there are w(1) liters burned.

As in one week there are 7 days, we can replace the r, that is a week, by something that represents 7 days. As 1 day is represented by t, one week can be 7t (in other words r = 7t). So, we have that the liters burned in one week are:

w(r) = w[7f(t)]

So, we represented the liters in one week by it measure of days.

So, we can post that the number of liters burned in 7 days is the same as the number of liters burned 1 day multiplied by 7 times. So:

w (r) = w[7 f(t)] = 7 f(t)

Here we hace the w function represented in terms of t instead of r.

3 0
2 years ago
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