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slega [8]
3 years ago
9

Plz use the graph help asap show work

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
7 0
Question 9.)
5-2(5)-1
10-1
Y=9
(X,Y) = 5,9
2- 2(2)-1
4-1
Y=3
(X,Y)= 2,3
0- 2(0)-1
2-1
Y=1
(X,Y)= 0,1
10)
4- 3(4)-7
12-7
Y=5
(X,Y)=4,3
1- 3(1)-7
3-7
Y=-4
(X,Y)= 1,-4
0- 3(0)-7
0-7
Y=-7
(X,Y)= 0,-7
(I believe you are supposed to plot the (X,Y) points on the graph)
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If Javier has a car with a mass of 250,000 kg and it is acted upon with a force of 300 Newtons. What is the acceleration?
natima [27]

Answer:

<u>The acceleration is 0.0012 meters/seconds²</u>

Step-by-step explanation:

1. Let's review the information given to answer the question correctly:

Mass = 250,000 kg

Force = 300 Newtons

2. What is the acceleration?

Based on Newton's Second Law of Motion, let's recall that:

Force = Mass * Acceleration

Acceleration = Force/Mass

Replacing with the real values, we have:

Acceleration = 300/250,0000

Acceleration = 0.0012 meters/seconds² (we can't express acceleration using Newtons/kilograms as the unit).

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Find the quotient. (2x^4-3x^3-6x^2+11x+8)/(x-2)
stiv31 [10]

Since the remander is 0 the answer is 2x²+5x+2

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The temperature steadily dropped 20 degrees over a period of 2.5 hours. What was the hourly temperature change?
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Im guessing you can make a ratio out of this, 20/2.5=x/1, 20=2.5x, x=8, so there was a degree change of 8 in one hour 
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3 years ago
WILL GIVE BRAINLIEST!!!!!
dimaraw [331]

1. f(x)=x²+10x+16

Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(As, a>0 the parabola is open upward), b=10. by putting the values.

-b/2a = -10/2(1) = -5

f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9

So, Vertex = (-5, -9)

Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).

Now find x-intercept put y=0 in the above equation. 0= x²+10x+16

x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2

From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.

2. f(x)=−(x−3)(x+1)

By multiplying the factors, the general form is f(x)= -x²+2x+3.

Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(As, a<0 the parabola is open downward), b=2. by putting the values.

-b/2a = -2/2(-1) = 1

f(-b/2a)= f(1)=-(1)²+2(1)+3= 4

So, Vertex = (1, 4)

Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).

Now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.

-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1

From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.

3. f(x)= −x²+4

Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(As, a<0 the parabola is open downward), b=0. by putting the values.

-b/2a = -0/2(-1) = 0

f(-b/2a)= f(0)= −(0)²+4 =4

So, Vertex = (0, 4)

Now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).

Now find x-intercept put y=0 in the above equation. 0= −x²+4

−x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2

From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.

4. f(x)=2x²+16x+30

Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(As, a>0 the parabola is open upward), b=16. by putting the values.

-b/2a = -16/2(2) = -4

f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2

So, Vertex = (-4, -2)

Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).

Now find x-intercept put y=0 in the above equation. 0=2x²+16x+30

2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3

From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.

5. y=(x+2)²+4

The general form of parabola is y=a(x-h)²+k , where vertex = (h,k)

if a>0 parabola is opened upward.

if a<0 parabola is opened downward.

Compare the given equation with general form of parabola.

-h=2 ⇒h=-2

k=4

so, vertex= (-2, 4)

As, a=1 which is greater than 0 so parabola is opened upward and the graph has minimum.

The graph is attached below.

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