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viktelen [127]
2 years ago
6

The graph of this system of equations is used to solve 4x^2-3x+6=2x^4-9x^3+2x What represents the solution set?

Mathematics
2 answers:
hammer [34]2 years ago
8 0

We are given equation : 4x^2-3x+6=2x^4-9x^3+2x.

Let us write it in form of a system of two equations:

y= 4x^2-3x+6 and

y =2x^4-9x^3+2x

Let us put those equations on a graphing calculator to get the graphs of the above system of equations.

<h3>From the graph, we can see that y-intercepts are : (0,0) and (0,6).</h3><h3>x-intercepts are (-0.45, 0) (0,0), (0.5,0) and (4.45,0)</h3><h3>y-coordinate of the intersection points : 14.69.</h3><h3>x-coordinates of the intersection points : -1.15</h3>

Alex17521 [72]2 years ago
8 0

Answer:

The solution is x-coordinate of the intersection points

D is correct

Step-by-step explanation:

Given: 4x^2-3x+6=2x^4-9x^3+2x

We need to find solution of given equation.

First we make system of equation and then find intersection point of graph.

y_1=4x^2-3x+6

y_2=2x^4-9x^3+2x

Now we draw the graph of system of equation using graphing calculator.

Please see the attachment for graph.

In graph both equation intersect at two points.

Point of intersection gives the solution of the equation.

x-coordinate of the intersection of graph gives the solution because function depends on x.

Hence, The solution is x-coordinate of the intersection points

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7 0
2 years ago
the scale from a square exercise mat in a gymnastic arena to a scale copy of the exercise mat is 8 m for every 2 in. The actual
m_a_m_a [10]

Answer: 18 and 36

Step-by-step explanation:

The scale is 8m for every 2 in and the actual area is 144m2 so we need to covert that into inches. You divide 144 by 8 which = 18 and then multiply 18 x 2 which is equal to 36.

7 0
2 years ago
A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks along the x-axis from the spotlight toward the bui
Sedbober [7]

Answer:

0.675 m/s

Step-by-step explanation:

Let height of shadow= y,CD=x

Height of man=2 m

Speed of man= \frac{dx}{dt}=1. 8 m/s

\triangle ABD\sim\triangle ECD

Therefore, \frac{AB}{EC}=\frac{BD}{CD}

\frac{y}{2}=\frac{12}{x}

xy=24

Differentiate w.r.t t

x\frac{dy}{dt}+y\frac{dx}{dt}=0

x\frac{dy}{dt}=-y\frac{dx}{dt}

\frac{dy}{dt}=-\frac{y}{x}\frac{dx}{dt}

When the man is 4 m from  the building

Then, we have x=12-4=8 m

\frac{dx}{dt}=1.8 m/s

Substitute the values in above equation then, we get

8y=24

y=\frac{24}{8}=3

Substitute the values then we get

\frac{dy}{dt}=-\frac{3}{8}\times 1.8=-0.675 m/s

Hence, the length of his shadow on the building decreasing at the rate 0.675 m/s.

8 0
2 years ago
Natalie tried to evaluate the expression (4-3.2-3)
shtirl [24]

Answer:

Natalie made a mistake from Step 1. The real answer is: \frac{1}{512}

Step-by-step explanation:

The correct evaluation of the expression is:

2^{-3}\times 4^{-3}

(2\times 4)^{-3}

8^{-3}

512^{-1}

\frac{1}{512}

4 0
2 years ago
A school gym is divided for a fair by bisecting its width and its length. Each half of the length is then​ bisected, forming 8 s
SVETLANKA909090 [29]

Answer:

Dimensions will be

New Length= x/16 units

New width = y/2 units

Area=xy/32 square units

Step-by-step explanation:

Lets assume the dimensions of the school gym to be ;

Length= x

width= y

Length and width are first bisected for a fair.This means by a half

Length will be =x/2

Width will be =y/2

Then the length is divided to form 8 sections that are equal, which will ;

(x/2)÷8

x/2 * 1/8 = x/16 units

New length = x/16 units in dimension

New width will be y/2 units in dimension

Area of each section will be;

length by width

(x/16 )*(y/2)

=xy/32 square units

7 0
2 years ago
Read 2 more answers
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