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gtnhenbr [62]
3 years ago
6

Given below is the even-number function. Which of the following are equal to E(12)? Check all that apply. E(n) = 2n A. 24 B. E(6

) + 6 C. E(8) + E(4) D. 14
Mathematics
1 answer:
Ray Of Light [21]3 years ago
4 0
Ummmm i believe your answer is A. just divide next time :)
You might be interested in
Need help asap please
Schach [20]
The answer is 7 meters.


Area is (pi)r^2
If we remove pi, we have 49=r^2
This means that r=7
4 0
3 years ago
Read 2 more answers
Ms. Ahmed has a rectangular garden where she grows tomatoes. The length of her garden is represented by t + 4 and the width by 2
Fantom [35]

Answer:

Area of the garden in standard form:

Area = (2t^2) + 7t - 4

The dimension of the garden is 8ft × 7ft

Step-by-step explanation:

Length of the rectangular garden, L = t+4

Width of the garden, B = 2t - 1

Area = length * width

Area = (t+4) * ( 2t-1)

Area = (2t^2) + 7t - 4..….......(1)

If the Area = 56 ft^2

Substitute Area into (1)

56 = (2t^2) + 7t - 4

(2t^2) + 7t - 60 = 0

(2t^2) - 8t + 15t - 60 = 0

2t(t - 4) + 15(t - 4) = 0

(2t+15) (t-4) = 0

Let 2t + 15 =0

t = -7.5

Let t - 4 = 0

t = 4

* When t = -7.5

Length, L = -7.5 + 4 = -3.5 ft

Width, B = 2(-7.5) -1 = -16 ft

Since length and breadth cannot be negative, t = -7.5 is not possible

** When t = 4

Length, L = 4+4 = 8 ft

Width, B = 2(4) - 1 = 7 ft

Therefore, the dimension of the garden is 8ft × 7ft

6 0
3 years ago
Read 2 more answers
What is 221,000,000,000,000,000,000 expressed in scientific notation? a.2.21 × 1021 b.2.21E20 c.2.21E21 d.2.21E-20 e.2.21 × 10-2
nekit [7.7K]

Answer:

221,000,000,000,000,000,000 = 2.21 x 10^20

therefore, the answer is b

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A metallurgist has an alloy with 10​% titanium and an alloy with 20​% titanium. He needs 100 grams of an alloy with 17​% titaniu
ser-zykov [4K]
\bf \begin{array}{lccclll}
&amount&concentration&
\begin{array}{llll}
concentration\\
amount
\end{array}\\
&-----&-------&-------\\
\textit{10\% alloy}&x&0.1&0.1x\\
\textit{20\% alloy}&y&0.2&0.2y\\
-----&-----&-------&-------\\
mixture&100&0.17&17
\end{array}

now, notice, we use the decimal format for the percent, namely 17% is 17/100 or 0.17 and so on

so....  we know, whatever "x" and "y" is, they must add up to 100 grams
thus    x + y = 100

and whatever the concentrated amount is, it must add up to 17 grams for the composition

thus   0.1x + 0.2y = 17

\bf \begin{cases}
x+y=100\implies \boxed{y}=100-x\\
0.1x+0.2y=17\\
----------\\
0.1x+0.2\left( \boxed{100-x} \right)=17
\end{cases}

solve for "x", to see how much is needed of the 10% alloy

what about "y"? well, y = 100 - x
4 0
3 years ago
11 times what give you 187
mr_godi [17]
17. 187 divided by 11 equals 17.
6 0
3 years ago
Read 2 more answers
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