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Tasya [4]
3 years ago
15

Which is an example of a quadratic function whose graph is entirely above the x-axis?

Mathematics
1 answer:
vredina [299]3 years ago
7 0
The answer is:  [D]:  " y = x² + 2x + 2 " .
______________________________________________
You might be interested in
Linar system in variables<br> -x-3y-2z=8,<br> -x+y+6z=, <br> x-9y-2z=4
docker41 [41]

Answer:

-x+y+6z=? You did not spcify so I just made it 0

x = -22/5, y = -4/5, z = -3/5

Step-by-step explanation:

Solve the following system:

{-x - 3 y - 2 z = 8 | (equation 1)

-x + y + 6 z = 0 | (equation 2)

x - 9 y - 2 z = 4 | (equation 3)

Subtract equation 1 from equation 2:

{-x - 3 y - 2 z = 8 | (equation 1)

0 x+4 y + 8 z = -8 | (equation 2)

x - 9 y - 2 z = 4 | (equation 3)

Divide equation 2 by 4:

{-x - 3 y - 2 z = 8 | (equation 1)

0 x+y + 2 z = -2 | (equation 2)

x - 9 y - 2 z = 4 | (equation 3)

Add equation 1 to equation 3:

{-x - 3 y - 2 z = 8 | (equation 1)

0 x+y + 2 z = -2 | (equation 2)

0 x - 12 y - 4 z = 12 | (equation 3)

Divide equation 3 by 4:

{-x - 3 y - 2 z = 8 | (equation 1)

0 x+y + 2 z = -2 | (equation 2)

0 x - 3 y - z = 3 | (equation 3)

Swap equation 2 with equation 3:

{-x - 3 y - 2 z = 8 | (equation 1)

0 x - 3 y - z = 3 | (equation 2)

0 x+y + 2 z = -2 | (equation 3)

Add 1/3 × (equation 2) to equation 3:

{-x - 3 y - 2 z = 8 | (equation 1)

0 x - 3 y - z = 3 | (equation 2)

0 x+0 y+(5 z)/3 = -1 | (equation 3)

Multiply equation 3 by 3:

{-x - 3 y - 2 z = 8 | (equation 1)

0 x - 3 y - z = 3 | (equation 2)

0 x+0 y+5 z = -3 | (equation 3)

Divide equation 3 by 5:

{-x - 3 y - 2 z = 8 | (equation 1)

0 x - 3 y - z = 3 | (equation 2)

0 x+0 y+z = -3/5 | (equation 3)

Add equation 3 to equation 2:

{-x - 3 y - 2 z = 8 | (equation 1)

0 x - 3 y+0 z = 12/5 | (equation 2)

0 x+0 y+z = -3/5 | (equation 3)

Divide equation 2 by -3:

{-x - 3 y - 2 z = 8 | (equation 1)

0 x+y+0 z = -4/5 | (equation 2)

0 x+0 y+z = -3/5 | (equation 3)

Add 3 × (equation 2) to equation 1:

{-x + 0 y - 2 z = 28/5 | (equation 1)

0 x+y+0 z = -4/5 | (equation 2)

0 x+0 y+z = -3/5 | (equation 3)

Add 2 × (equation 3) to equation 1:

{-x+0 y+0 z = 22/5 | (equation 1)

0 x+y+0 z = -4/5 | (equation 2)

0 x+0 y+z = -3/5 | (equation 3)

Multiply equation 1 by -1:

{x+0 y+0 z = -22/5 | (equation 1)

0 x+y+0 z = -4/5 | (equation 2)

0 x+0 y+z = -3/5 | (equation 3)

Collect results:

Answer: {x = -22/5, y = -4/5, z = -3/5

8 0
3 years ago
Which expression is the product of (x + 4i)(x − 4i)(x + 4)?
Sophie [7]
Answer= x³+4x²+16x+64

Expand the following:(x + 4 i) (x - 4 i) (x + 4)
(x - 4 i) (x + 4) = (x) (x) + (x) (4) + (-4 i) (x) + (-4 i) (4) = x^2 + 4 x - 4 i x - 16 i = -16 i + (4 - 4 i) x + x^2:
-16 i + (-4 i + 4) x + x^2 (4 i + x)
 | | | | x | + | 4 i
 | | x^2 | + | (4 - 4 i) x | - | 16 i
 | | | | (-16 i) x | + | 64
 | | (4 - 4 i) x^2 | + | (16 + 16 i) x | + | 0
x^3 | + | (4 i) x^2 | + | 0 | + | 0
x^3 | + | 4 x^2 | + | 16 x | + | 64:
Answer: x^3 + 4 x^2 + 16 x + 64
7 0
3 years ago
Read 2 more answers
Metal A has a density of 12.7 g/cm3.
stira [4]

Using the given information, the density of metal B is 8.58 g/cm³

<h3>Calculating density </h3>

From the question, we are to determine the density of metal B

From the given information,

55 g of metal A is combined with metal B to create an alloy with mass 90 g

∴ Mass of metal B in the alloy = 90 g - 55 g = 35 g

Now, we will determine the volume of metal A in the alloy

Using the formula,

Density = Mass / Volume

∴ Volume = Mass / Density

Volume of metal A in the alloy = 55/12.7

Volume of metal A in the alloy = 4.33 cm³

Then, we will determine the volume of the alloy

Volume of the alloy = 90/10.7

Volume of the alloy = 8.41 cm³

∴ Volume of metal B in the alloy = 8.41 cm³- 4.33 cm³

Volume of metal B in the alloy = 4.08 cm³

Now, for the density of metal B

Density = mass / volume

Density of metal B = 35 / 4.08

Density of metal B = 8.58 g/cm³

Hence, the density of metal B is 8.58 g/cm³.

Learn more on Calculating density here: brainly.com/question/13986137

#SPJ1

5 0
1 year ago
if it took 9 hours to mow 15 lawns, then at that rate, how many lawns could be mowed in 27 hours? What was the rate of lawns mow
sammy [17]

Answer:

Step-by-step explanation:

15lawns/9hours= 1.6 lawns mowed per hour

1.6lph*27hours=43.2 lawns

6 0
2 years ago
How many 5-millileter test tubes filled with water will it take to fill a 1 leter container?
aliya0001 [1]
One liter is equal to 1000 milliliters. Use the conversion to find your answer. 1000 milliliters / 5mL per tube = 200 tubes
5 0
3 years ago
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