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Vitek1552 [10]
2 years ago
7

She has 100 chairs that she needs to arrange in a perfect square. How many rows of chairs does she need to make?

Mathematics
2 answers:
Mice21 [21]2 years ago
8 0
She needs to make 10 rows of chairs !!
Nookie1986 [14]2 years ago
3 0

Answer:

She needs to make 10 rows of chairs

Hope this helps!!

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Help will mark Brainliest!!​
BartSMP [9]

Answer:

-20.5

Step-by-step explanation:

Using a basic calculator, you divide 10.25 by -.5 which easily gives -20.5

5 0
1 year ago
What is the mode of this data set? 64°, 72°, 74°, 68°, 70°, 68°, 68°, 75°, and 44
butalik [34]

Hey there!

Mode is basically the number you see more than once,

64, 72, 74, 68, 70, 68, 68, 75,  44\\ \\ Mode: 68

Your answer would be: 68 because it appeared approximately 3 times in this data set

Good luck on your assignment and enjoy your day!

~LoveYourselfFirst:)

6 0
3 years ago
Read 2 more answers
6
Ede4ka [16]

Answer:

5 percent per annum

Step-by-step explanation:

interest rate:

interest: $1800

principal: $18000

duration: 2 years

1800/ 18000 * 100

result is 10%

Since interest is 10% in 2 years, 10% / 2

interest is 5% per year.

8 0
2 years ago
PLEASE HELP ASAP
pishuonlain [190]
You have that:

 1. By definition, a binomial is a polynomial formed by two terms. 
 2. A quintic binomial is a binomial of degree 5. This means that its highest exponent is 5.
 3. Keeping this on mind, you must identify which binomial has degree 5.
 4. As you can see, the expression <span>3x^5+2 is a binomial, because it has two terms and the highest degree is 5.
 5. Thefore, the answer is: </span>3x^5+2
8 0
3 years ago
Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt
polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt

=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k

The first integral is trivial since (\tan t)'=\sec^2t.

The second can be done by substituting u=t^2-1:

u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C

The third can be found by integrating by parts:

u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t

\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8

\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C

8 0
3 years ago
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