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tamaranim1 [39]
2 years ago
10

If the sides of a triangle have the following lengths, find a range possible values for x

Mathematics
1 answer:
makvit [3.9K]2 years ago
4 0

Answer:

9.5 < x < 15

Step-by-step explanation:

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Evaluate the expression : common log of square root 9/25.
tekilochka [14]

You want log √(9/25).  Recognizing that √9 = 3 and that √25 = 5, we get

log 3/5, which by rules of logs comes out to log 3 - log 5.

To four decimal places:

log 3 - log 5 = 0.4771 - 0.6990, or -0.2218.

4 0
3 years ago
Midpoint of the segment between the points (−5,13) and (6,4)
Leokris [45]

Answer: (0.5,8.5)

Step-by-step explanation:

We need to use the following formula to find the Midpoint "M":

M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Given the points (-5,13) and (6,4) can identify that:

x_1=-5\\x_2=6\\\\y_1=13\\y_2=4

The final step is to substitute values into the formula.

Therefore, the midpoint of the segment between the points (-5,13) and (6,4) is:

M=(\frac{-5+6}{2},\frac{13+4}{2})\\\\M=(\frac{1}{2},\frac{17}{2})\\\\M=(0.5,8.5)

6 0
3 years ago
Read 2 more answers
[challenge] you are hoping to receive a scholarship for volleyball and are entering your personal information and statistics on
andriy [413]
Height without rounding - 5.89

5.89

Rounded - 5.9 feet tall
7 0
3 years ago
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WHAT IS THE REMAINDER WHEN <img src="https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%20%7D" id="TexFormula1" title="32^{37^{32} }"
Feliz [49]

Recall Euler's theorem: if \gcd(a,n) = 1, then

a^{\phi(n)} \equiv 1 \pmod n

where \phi is Euler's totient function.

We have \gcd(9,32) = 1 - in fact, \gcd(9,32^k)=1 for any k\in\Bbb N since 9=3^2 and 32=2^5 share no common divisors - as well as \phi(9) = 6.

Now,

37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} =  32\cdot\left(32^{(\cdots)}\right)^6

where the c_i are positive integer coefficients from the binomial expansion. By Euler's theorem,

\left(32^{(\cdots)\right)^6 \equiv 1 \pmod9

so that

32^{37^{32}} \equiv 32\cdot1 \equiv \boxed{5} \pmod9

7 0
2 years ago
If 1200 cm^2 of material is available to make a box with a square base and an open top find the largest possible volume of the b
strojnjashka [21]
Surface area of box=1200 cm² 
<span>Volume of box=s²h </span>
<span>s = side of square base </span>
<span>h = height of box </span>
<span>S.A. = s² + 4sh </span>
<span>S.A. = surface area or 1200 cm², s²
 = the square base, and 4sh = the four 'walls' of the box. </span>
<span>1200 = s² + 4sh </span>
<span>1200 - s² = 4sh </span>
<span>(1200 - s²)/(4s) = h </span>


<span>v(s) = s²((1200 - s²)/(4s)) </span>
<span>v(s) = s(1200 - s²)/4 . </span>
<span>v(s) = 300s - (1/4)s^3</span>

by derivating
<span>v'(s) = 300 - (3/4)s² </span>
<span>0 = 300 - (3/4)s² </span>
<span>-300 = (-3/4)s² </span>
<span>400 = s² </span>
<span>s = -20 and 20. </span>
again derivating
<span>v"(s) = -(3/2)s </span>
<span>v"(-20) = -(3/2)(-20) </span>
<span>v"(-20) = 30 </span>
<span>v"(20) = -(3/2)(20) </span>
<span>v"(20) = -30 </span>


<span>v(s) = 300s - (1/4)s^3 </span>
<span>v(s) = 300(20) - (1/4)(20)^3 </span>
<span>v(s) = 6000 - (1/4)(8000) </span>
<span>v = 6000 - 2000 
v=4000</span>
4 0
3 years ago
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