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Talja [164]
3 years ago
5

First correct answer gets best marks ​

Mathematics
2 answers:
Mandarinka [93]3 years ago
6 0

Answer:

the answer would be x is less than 6.

Step-by-step explanation:

the reason why it would not be x is less than or equal to 6 is that the circle is not filled in.

Zina [86]3 years ago
4 0

Answer:

B

Step-by-step explanation:

x≤6

We can see from the graph that it starts from 6 and goes to 5, 4, 3, 2.

Hope this helps ;) ❤❤❤

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Suppose you start at the origin, move along the x-axis a distance of 4 units in the positive direction, and then move downward a
Zanzabum

Answer:

(4,0,-5)

Step-by-step explanation:

Since we move 4 points along +ve x-axis so,  x coordinate is 4

We did not move any point along y-axis so  y coordinate is 0

we moved 5 points along -ve z-axis so z coordinate is -5

so our position is (x,y,z) = (4,0,-5)

3 0
3 years ago
Help pls will give brainliest to correct answer
Marysya12 [62]
D is correct answers
7 0
2 years ago
If a 100.8 kg cement bag had 20.35 kg removed from it, what is the new weight of the bag?
Snowcat [4.5K]
Just subtract 20.35 from 100.8
100.8 \: kg - 20.35 \: kg = 80.45 \: kg

if your bag weighs 100.8 kg and you scoop out 20.35 kg then weigh your bag it will be 80.45 bc u took out 20.35 kg of the cement to use. hope that explains it
8 0
3 years ago
Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)
masha68 [24]

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

y=ax^2+bx+c            ...(i)

It is passes through the point (0,11). So, substitute x=0,y=11 in (i).

11=a(0)^2+b(0)+c

11=c

Putting c=11 in (i), we get

y=ax^2+bx+11               ...(ii)

The quadratic function passes through the point (5,31). So, substitute x=5,y=31 in (ii).

31=a(5)^2+b(5)+11

31-11=a(25)+5b

20=25a+5b

Divide both sides by 5.

4=5a+b                  ...(iii)

The quadratic function passes through the point (3,11). So, substitute x=3,y=11 in (ii).

11=a(3)^2+b(3)+11

11-11=a(9)+3b

0=9a+3b

Divide both sides by 3.

0=3a+b                 ...(iv)

Subtracting (iv) from (iii), we get

4-0=5a+b-3a-b

4=2a

\dfrac{4}{2}=a

2=a

Putting a=2 in (iv), we get

0=3(2)+b

0=6+b

-6=b

Putting a=2,b=-6 in (ii), we get

y=(2)x^2+(-6)x+11

y=2x^2-6x+11

Therefore, the required quadratic equation is y=2x^2-6x+11.

7 0
2 years ago
Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
hammer [34]
Do u still need help?
6 0
2 years ago
Read 2 more answers
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