Variable y varies directly with variable x, and y = 6 when x = 4. Enter the constant of variation in lowest terms.

y =__x

Answers

Answer 1

Answer: You need to plug K in and solve for K.
6=K4
6/4=K
you can reduce to 3/2=K

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Find the possible values for s in the inequality 12s – 20 ≤ 50 – 3s – 25.

Answers

12s - 20 < 50 - 3s - 25

12s - 20 < 50 - 3s - 25
+3s +3s
15s - 20 < 50 - 25
+ 20 + 20
15s < 70 - 25

15s < 45
÷ 15 ÷15
s < 3

The value of s is less than or equal to 3. So s can be 3, 2, 1, 0, and negative numbers.

s = 3 ⇒ 12(3) - 20 < 50 - 3(3) - 25 ; 36 - 20 < 50 - 9 - 25 ; 16 < 16
s = 2 ⇒ 12(2) - 20 < 50 - 3(2) - 25 ; 24 - 20 < 50 - 6 - 25 ; 4 < 19

Solve for the variable-6a+3=-3(2a-1)

Can some one help me. But explain in step by step?

Answers

Step 1: -6a + 3 = -3(2a - 1)
Step 2: -6a + 3 = -3(2a) + 3(1)
Step 3: -6a + 3 = -6a + 3
Step 4:+6a +6a
Step 5: 3 = 3

The variable of x is equal to Any infinite number.

The two cylinders are similar. If the ratio of their surface areas is 9/1.44, find the volume of each cylinder. Round your answer to the nearest hundredth. A.small cylinder: 152.00 m3
large cylinder: 950.02 m3
B.
small cylinder: 972.14 m3
large cylinder: 12,924.24 m3
C.
small cylinder: 851.22 m3
large cylinder: 13,300.25 m3
D.
small cylinder: 682.95 m3
large cylinder: 13,539.68 m3

Answers

r1² π : r2² π = 9 : 1.44
r1 : r 2 = 3 : 1.2
r2 = 1.2 · r1 / 3
r 2 = 0.4 · r1
If the two cylinders are similar than also: h 2 = 0.4 · h 2
V 1 = r1² · π · h1
V 2 = ( 0.4 r1)² π · 0.4 · h1 = 0.064 r1² · π · h1
V 2 ( small cylinder )= 0.064 · V 1 ( large cylinder )
851.22 m³= 13.300.25 m³ · 0.064
Answer: C )

The two cylinders are similar. If the ratio of their surface areas is 9/1.44, find the volume of each cylinder. Round your answer to the nearest hundredth.

Answer: out of all the available options the one that shows the correct volumes for each cylinder is answer choice C) small cylinder: 851.22 m³ & large cylinder: 13,300.25 m³.

I hope it helps, Regards.

If -nx2 + tx + c = 0, what is x equal to?

Answers

$f(x)=-nx^2+tx+c=0\n\n \Delta=t^2-4\cdot(-n)\cdot c=t^2+4cn\n\n 1.\n \Delta<0 \Rightarrow x\in\emptyset\n\n 2.\n \Delta=0\n x=(-t)/(2\cdot(-n))=(t)/(2n)\n\n 3.\n \Delta>0\n √(\Delta)=√(t^2+4cn)\n x_1=(-t-√(t^2+4cn))/(2\cdot(-n))=(t+√(t^2+4cn))/(2n)\n x_2=(-t+√(t^2+4cn))/(2\cdot(-n))=(t-√(t^2+4cn))/(2n)\n$

Quadratic formula 2x²+9x-4=0

Answers

$2x^2+9x-4=0 \n \na=2 , b =9 , \ c=- 4 \n \n\Delta = b^(2)-4ac = 9^(2)-4*2* (- 4)= 81+32 =113 \n \nx_(1)=(-b-√(\Delta ))/(2a) =(-9- √(113))/(4) \n\nx_(2)=(-b+√(\Delta ))/(2a) = (-9+√(113))/(4)$

$2x^2+9x-4=0\ \ \ \ /:2\n\nx^2+(9)/(2)x-2=0\n\nx^2+2x\cdot(9)/(4)+((9)/(4))^2-((9)/(4))^2=2\n\n(x+(9)/(4))^2-(81)/(16)=2\n\n(x+(9)/(4))^2=(32)/(16)+(81)/(16)\n\n(x+(9)/(4))^2=(113)/(16)\iff x+(9)/(4)=-\sqrt(113)/(16)\ \vee\ x+(9)/(4)=\sqrt(113)/(16)\n\nx=-(9)/(4)-(√(113))/(4)\ \vee\ x=-(9)/(4)+(√(113))/(4)\n\nx=-(9+√(113))/(4)\ \vee\ x=-(9-√(113))/(4)$