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Vitek1552 [10]
2 years ago
14

5. The hypotenuse of a right triangle is 14 in. If the base of the triangle is 2 inches determine the

Mathematics
2 answers:
kifflom [539]2 years ago
5 0

Answer:

13.8564...

Step-by-step explanation:

pythagorean theorem

Shtirlitz [24]2 years ago
4 0

Answer:

13.9642400438 inches

Step-by-step explanation:

2^2+b^2=14^2

4+b^2=196

b^2=195

b=13.9642400438

Therefore, 13.9642400438 inches

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crimeas [40]

Answer:

  c.  3a^2b^11/2

Step-by-step explanation:

The applicable rule of exponents is ...

  (a^b)/(a^c) = a^(b-c)

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 \dfrac{-18a^{-2}b^5}{-12a^{-4}b^{-6}}=\dfrac{-18}{-12}a^{-2-(-4)}b^{5-(-6)}=\boxed{\dfrac{3a^2b^{11}}{2}}

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2 years ago
(30 POINTS) The 20% off sale is a better deal than the $200 rebate or $150 coupon for the $1,500 dining set. The Porters budgete
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<span>Yes, the Porters are under budget. The 20% off sale is the best deal because $1,500 times 0.20 is $300. $300 is a bigger discount than the $150 coupon and $200 rebate. $300 off $1,500 is $1,200, which is less than the amount budgeted.</span>
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2 years ago
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(3x+2)∧2=9 how do i solve this in the square root property
Irina18 [472]
(3x+2)^2=9 \\\\9x^2+12x+4-9=0\\\\9x^2+12x-5=0\\\\a=9,\ \ b=12, \ \ c=-5

x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-12-\sqrt{12^2-4 \cdot9 \cdot (-5)}}{2 \cdot 9}=\frac{-12-\sqrt{144+180}}{18}=\\\\=\frac{-12-\sqrt{324}}{18}=\frac{-12-18}{18}=\frac{-30}{18}=-\frac{5}{3}\\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-12+\sqrt{12^2-4 \cdot9 \cdot (-5)}}{2 \cdot 9}=\frac{-12+18}{18}=\frac{6}{18}=\frac{1}{3} \\\\Answer: \ x=-\frac{5}{3}\ \ and \ \ x=\frac{1}{3}
 

6 0
3 years ago
Find \tan\left(\frac{17\pi}{12}\right)tan( 12 17π ​ )tangent, left parenthesis, start fraction, 17, pi, divided by, 12, end frac
uranmaximum [27]

One way to do this is to notice

\dfrac{17\pi}{12}=\dfrac\pi6+\dfrac{5\pi}4

Then

\tan\dfrac{17\pi}{12}=\tan\left(\dfrac\pi6+\dfrac{5\pi}4\right)=\dfrac{\tan\frac\pi6+\tan\frac{5\pi}4}{1-\tan\frac\pi6\tan\frac{5\pi}4}

We have

\tan\dfrac\pi6=\dfrac{\sin\frac\pi6}{\cos\frac\pi6}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}

and since \tan x has a period of \pi,

\tan\dfrac{5\pi}4=\tan\left(\pi+\dfrac\pi4\right)=\tan\dfrac\pi4=1

and so

\tan\dfrac{17\pi}{12}=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3

7 0
3 years ago
on a piece of paper, graph this system of inequalities. Then determine which region contains the solution to the system. a pictu
slava [35]

C. Region D

Step-by-step explanation:

Just plug the equations into desmos the calculator will graph it and show you with colors where the solution is.

4 0
3 years ago
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