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pishuonlain [190]
2 years ago
13

Solve for x in the following equation.

Mathematics
1 answer:
KonstantinChe [14]2 years ago
5 0

Hello,

3 {}^{a}  = 3 {}^{b} \Leftrightarrow \: a = b

3 {}^{x}  =  {3}^{2} \Leftrightarrow \: x = 2

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Help me please . quick :)
ycow [4]

Answer:

x= 1

QS = 10 cm

QRS = 40cm

Step-by-step explanation:

4x + 1 = 5x

x= 1 cm

QS = 4x + 1 + 5x

= 9x + 1

= 9 + 1

= 10cm

QRS perimeter

QS + RQ + RS but ( RQ = RS)

= 10 + 15 + 15

= 40cm

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2 years ago
The ordered pair (-4, -5) is a solution of the following system of equations 2x+y=13 19x+13y=15
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Answer:

Step-by-step explanation

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There are two major tests of readiness for college, the act and the sat. act scores are reported on a scale from 1 to 36. the di
ch4aika [34]
Standardized z score for the act result  =  (30 - 21.5) / 5.4  = 1.574

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3 years ago
Find the equation of a sphere if one of its diameters has endpoints: (-9, -12, -6) and (11, 8, 14).
baherus [9]

Answer:

Hence, the equation of a sphere with one of its diameters with endpoints (-9, -12, -6) and (11, 8, 14) is (x-1)^{2}+(y+2)^{2}+(z-4)^{2} = 30.

Step-by-step explanation:

There are two kew parameters for a sphere: Center (h, k, s) and Radius (r). The radius is the midpoint of the line segment between endpoints. That is:

C(x,y,z) = \left(\frac{-9+11}{2},\frac{-12+8}{2},\frac{-6+14}{2}   \right)

C(x,y,z) = (1,-2,4)

The radius can be found by halving the length of diameter, which can be determined by knowning location of endpoints and using Pythagorean Theorem:

r = \frac{1}{2}\cdot \sqrt{(-9-11)^{2}+(-12-8)^{2}+(-6-14)^{2}}

r = 10\sqrt{3}

The general formula of a sphere centered at (h, k, s) and with a radius r is:

(x-h)^{2}+(y-k)^{2}+(z-s)^{2} = r^{2}

Hence, the equation of a sphere with one of its diameters with endpoints (-9, -12, -6) and (11, 8, 14) is (x-1)^{2}+(y+2)^{2}+(z-4)^{2} = 30.

8 0
3 years ago
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