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ad-work [718]
2 years ago
6

PLEASE I NEED HELPPP !!!!! please help

Mathematics
2 answers:
weeeeeb [17]2 years ago
7 0
Yeah we have to see it properly to give you the right answer
Strike441 [17]2 years ago
3 0
Can you take a better picture because it’s hard to see where the line marks. or tell me the coordinates.
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Write seventy two thousand and thirty two ?
miv72 [106K]
In numbers, <span>seventy two thousand and thirty two is 72032</span>
8 0
3 years ago
Landon played football last week he played 22 minutes on Monday 19 minutes on Tuesday and 33 minutes on Wednesday on Thursday he
ladessa [460]

Answer:

32

Step-by-step explanation:

M: 22

T:19

W:33

Th: 2m, if m=22, Th=44

22+19+33+44=118

150-118=32

So he played for 32 minutes on Friday

4 0
3 years ago
Help me please!! Thank you!!!
kotykmax [81]

Answer:

the first one and the last one

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Can a line and a point be collinear
xxTIMURxx [149]

Answer:

Points that lie on the same line are called collinear points. If there is no line on which all of the points lie, then they are noncollinear points.

Step-by-step explanation:

I dont have an explanation

3 0
3 years ago
Consider the following quadratic function Part 3 of 6: Find the x-intercepts. Express it in ordered pairs.Part 4 of 6: Find the
maksim [4K]

Answer:

The line of symmetry is x = -3

Explanation:

Given a quadratic function, we know that the graph is a parabola. The general form of a parabola is:

y=ax^2+bx+c

The line of symmetry coincides with the x-axis of the vertex. To find the x-coordinate of the vertex, we can use the formula:

x_v=-\frac{b}{2a}

In this problem, we have:

y=-x^2-6x-13

Then:

a = -1

b = -6

We write now:

x_v=-\frac{-6}{2(-1)}=-\frac{-6}{-2}=-\frac{6}{2}=-3

Part 3:

For this part, we need to find the x-intercepts. This is, when y = 0:

-x^2-6x-13=0

To solve this, we can use the quadratic formula:

x_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot(-1)\cdot(-13)}}{2(-1)}

And solve:

x_{1,2}=\frac{6\pm\sqrt{36-52}}{-2}x_{1,2}=\frac{-6\pm\sqrt{-16}}{2}

Since there is no solution to the square root of a negative number, the function does not have any x-intercept. The correct option is ZERO x-intercepts.

Part 4:

To find the y intercept, we need to find the value of y when x = 0:

y=-0^2-6\cdot0-13=-13

The y-intercept is at (0, -13)

Part 5:

Now we need to find two points in the parabola. Let-s evaluate the function when x = 1 and x = -1:

x=1\Rightarrow y=-1^2-6\cdot1-13=-1-6-13=-20x=-1\Rightarrow y=-(-1)^2-6\cdot(-1)-13=-1+6-13=-8

The two points are:

(1, -20)

(-1, -8)

Part 6:

Now, we can use 3 points to find the graph of the parabola.

We can locate (1, -20) and (-1, -8)

The third could be the vertex. We need to find the y-coordinate of the vertex. We know that the x-coordinate of the vertex is x = -3

Then, y-coordinate of the vertex is:

y=-(-3)^2-6(-3)-13=-9+18-13=-4

The third point we can use is (-3, -4)

Now we can locate them in the cartesian plane:

And that's enough to get the full graph:

8 0
1 year ago
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