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ser-zykov [4K]
3 years ago
14

1. _____ any push or pull on an object

Mathematics
1 answer:
borishaifa [10]3 years ago
5 0
1. Force any push or pull on an object.             2. Displacement is an object changing position over time.
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What is the domain and range of <br> f(x)=2x^2+6x+2
lbvjy [14]

As is the case for any polynomial, the domain of this one is (-infinity,  +infinity).

To find the range, we need to determine the minimum value that f(x) can have.  The coefficients here are a=2, b=6 and c = 2,

The x-coordinate of the vertex is  x = -b/(2a), which here is x = -6/4 = -3/2.

Evaluate the function at x = 3/2 to find the y-coordinate of the vertex, which is also the smallest value the function can take on.  That happens to be y = -5/2, so the range is [-5/2, infinity).

3 0
3 years ago
Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.
Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ;  x = \frac{2-(3)\sqrt{2}}{2}

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

for the equation ax² + bx + c = 0

thus,

the roots are

x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}

or

x = \frac{6\pm\sqrt{36-12}}{2}

or

x = \frac{6+\sqrt{24}}{2} and, x = x = \frac{6-\sqrt{24}}{2}

or

x = \frac{6+2\sqrt{6}}{2} and, x = x = \frac{6-2\sqrt{6}}{2}

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}

or

x = \frac{4\pm\sqrt{16+56}}{4}

or

x = \frac{4+\sqrt{72}}{4} and, x = x = \frac{4-\sqrt{72}}{4}

or

x = \frac{4+\sqrt{2^2\times3^2\times2}}{2} and, x = x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}

or

x = \frac{4+(2\times3)\sqrt{2}}{2} and, x = x = \frac{4-(2\times3)\sqrt{2}}{4}

or

x = \frac{2+3\sqrt{2}}{2} and, x = \frac{2-(3)\sqrt{2}}{2}

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ; x = \frac{2-(3)\sqrt{2}}{2}

7 0
3 years ago
A farmer sells 8.9 kilograms of pears and apples at the farmer's market. 1 4 of this weight is pears, and the rest is apples. Ho
dmitriy555 [2]

Answer:

7.5 kilograms of apples

Step-by-step explanation:

Take the total amount and subtract the amount of pears.

8.9-1.4=7.5

4 0
2 years ago
Read 2 more answers
When solving negative -1/3.(x − 15) = −4, what is the correct sequence of operations
FrozenT [24]

Answer:

x=27

Step-by-step explanation:

i would give you the steps but honestly, im tired of giving 20 steps for every problem i answer.

5 0
3 years ago
Select the pair of consecutive integers between which a real zero of f(x) = x^3 + 5x^2 –x – 6 exists.
Svetllana [295]
<span>f(x)=0 at 
rounded up 
-4.96 
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1.1</span><span>
</span>
5 0
3 years ago
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