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Ronch [10]
2 years ago
12

HELP ME PLEASE !! I need help with this graph ... will mark brainliest !!

Mathematics
2 answers:
Bingel [31]2 years ago
5 0

Answer:

its y=-2(x+1)^2+5

Step-by-step explanation:

postnew [5]2 years ago
5 0

Answer:

hello

Step-by-step explanation:

start new meeting please I will join

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Fifteen more than half a number is 9
rusak2 [61]

For this case, the first thing we must do is define a variable.

We have then:

x: unknown number

We now write the equation that models the problem:

(\frac{1}{2}) x + 15 = 9

From here, we clear the value of x.

We multiply both sides of the equation by 2:

(\frac{2}{2}) x + 2 (15) = 2 (9)\\x + 30 = 18

We subtract 30 on both sides of the equation:

x + 30 - 30 = 18 - 30\\x = -12

Answer:

The value of the unknown number is given by:

x = -12

6 0
3 years ago
Read 2 more answers
27.64x3 estimate product
ki77a [65]
If you round up the 27.64 to 28 and then multiply it bye 3 it eaquals 84


6 0
3 years ago
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A ball is thrown up in the air, and it's height (in feet) as a function of time (in seconds) can be written as h(t) = -16t2 + 32
olga2289 [7]

Answer:

1) Using properties of the quadratic equation:

Here the height equation is:

h(t) = -16t^2 + 32t + 6

We can see that the leading coefficient is negative, this means that the arms of the graph will open downwards.

Then, the vertex of the quadratic equation will be the maximum.

Remember that for a general quadratic equation:

a*x^2 + b*x + c = y

the x-value of the vertex is:

x = -b/2a

Then in our case, the vertex is at:

t = -32/(2*-16) = 1

The maximum height will be the height equation evaluated in this time:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

2) Second method, using physics.

We know that an object reaches its maximum height when the velocity is equal to zero (the velocity equal to zero means that, at this point, the object stops going upwards).

If the height equation is:

h(t) = -16*t^2 + 32*t + 6

the velocity equation is the first derivation of h(t)

Remember that for a function f(x) = a*x^n

we have that:

df(x)/dx = n*a*x^(n-1)

Then:

v(t) = dh(t)/dt = 2*(-16)*t + 32 + 0

v(t) = -32*t + 32

Now we need to find the value of t such that the velocity is equal to zero:

v(t) = 0 = -32*t + 32

       32*t = 32

            t = 32/32 = 1

So the maximum height is at t = 1

(same as before)

Now we just need to evaluate the height equation in t = 1:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

4 0
3 years ago
Explain why you would make one of the addends a tens number when solving an addition problem
Lyrx [107]
It is showing you how to be
3 0
3 years ago
755,082 rounded to the nearest 10,000
Tems11 [23]
760,000 is the answer
4 0
3 years ago
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