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erastovalidia [21]
3 years ago
13

Find the value of x. Please please help!!

Mathematics
1 answer:
Luden [163]3 years ago
5 0

Answer:

i have no ideia

Step-by-step explanation:

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Hello! who are you tell me ?

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2 years ago
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What is the domain of the function f(x) = 2/5√x ?
lisov135 [29]

Answer:

D: All real numbers greater than or equal to 0.

Step-by-step explanation:

6 0
3 years ago
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Can someone help me with these
Stolb23 [73]

Answer:

Problem 2) : the gradient is "-2", and the y-intercept is "3"

Problem 3)

A is  y=2x+2

B is y=x+2  

Step-by-step explanation:

Problem 2)

In the line given by the equation:   y=-2x+3

the "gradient" (also known as "slope") is the numerical coefficient that multiplies the variable "x". So in this case the gradient is "-2"

the y-intercept is the numerical term "+3" because that is the y-value result of evaluating the expression for x = 0

y=-2x+3\\y=-2\,(0)+3\\y=3

Problem 3)

Consider the two lines :

y=x+2  and  y=2x+2

notice that both have the same y-intercept (that is the numerical term "2" at the end of both expressions. That means that both lines cross the y-axis at the point y=2.

Now notice that the gradient of one of them is "1"  (for y=x+2 ) that is the coefficient that multiplies the variable "x". While for the other line ( y=2x+2)  the gradient is "2" and therefore steeper than the previous one.

Then, the line identified as "A" which is the one with steeper gradient, corresponds to the equation   y=2x+2, and the line identified with "B" is the one with smaller gradient  y=x+2  .

6 0
3 years ago
Find the missing factor in the exponential form 60^9=12^9*(Blank)
skad [1K]
(60)^9 = (12)^9 × (5)^9
6 0
3 years ago
A ball is launched straight up in the air from a height of 6 feet. Its velocity (feet/second) t seconds after launch is given by
german

Step-by-step explanation:

A ball is launched straight up in the air from a height of 6 feet. The velocity as a function of time t is given by :

f(t) = -32 t+285

Height of the ball is :

h(t)=\int\limits{f(t){\cdot} dt}\\\\h(t)=\int\limits{(-32t+285){\cdot} dt}\\\\h(t)=-16t^2+285t+C

C is constant. Here the ball is launched from a height of 6 feet. So,

h(t)=-16t^2+285t+6

At t = 2 s, h(t)=-16(2)^2+285(2)+6=512\ m

At t = 9 s, h(t)=-16(9)^2+285(9)+6=1275\ m

Between 2 s and 9 s, the ball's height changed is : 1275 - 512 = 763 m.

6 0
3 years ago
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