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azamat
3 years ago
6

What is the range of the function graphed below?

Mathematics
1 answer:
Mumz [18]3 years ago
4 0

Answer:

-\infty < y < -2

Step-by-step explanation:

Given

The attached graph

Required

The range of the graph

From the graph we can see that the curve points down towards the y-axis,

This means that y starts from a -\infty number

Coming up, the curve stops increasing at 2 and becomes horizontal; This means that y ends at 2

<em>Hence, the range is: </em>-\infty < y < -2<em></em>

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You sold 500 shares of Intel for $26.75 per share. Your broker charges you a
Natali [406]

Answer: $13339.25

Step-by-step explanation:

Since 500 shares of Intel were sold for $26.75 per share, the total amount will be:

= 500 × $26.75

= $13375

Since the broker charges a

commission of $2.75 per thousand dollars of stock bought or sold, the amount charged by the broker on $13375 will be:

= $2.75 × 13

= $35.75

The amount that'll then be credited into ones account will be:

= $13375 - $35.75

= $13339.25

8 0
3 years ago
5g + 7g and g(5 + 7) when g=6
oksian1 [2.3K]

Answer:

Value of equation = 144

Step-by-step explanation:

Given:

5g + 7g + g(5 + 7)

when g = 6

Find:

Value of equation

Computation:

5g + 7g + g(5 + 7)

5g + 7g + g(12)

5(6) + 7(6) + (6)(12)

30 + 42 + 72

Value of equation = 144

4 0
3 years ago
Math word problems please help
tatiyna
Answer: leave the math class
3 0
2 years ago
Rewrite each expression using exponents (-19)(-19)(-19)(-19)(-19)
Serggg [28]

Answer:

-19^5 or -19 to the fifth

4 0
4 years ago
A particle in the first quadrant is moving along a path described by the equation LaTeX: x^2+xy+2y^2=16x 2 + x y + 2 y 2 = 16 su
ElenaW [278]

Answer:

\frac{50}{3} cm/sec.

Step-by-step explanation:

We have been given that a particle in the first quadrant is moving along a path described by the equation x^2+xy+2y^2=16 such that at the moment its x-coordinate is 2, its y-coordinate is decreasing at a rate of 10 cm/sec. We are asked to find the rate at which x-coordinate is changing at that time.

First of all, we will find the y value, when x =2 by substituting x =2 in our given equation.

2^2+2y+2y^2=16

4-16+2y+2y^2=16-16

2y^2+2y-12=0

y^2+y-6=0

y^2+3y-2y-6=0  

(y+3)(y-2)=0

(y+3)=0,(y-2)=0

y=-3,y=2

Since the particle is moving in the 1st quadrant, so the value of y will be positive that is y=2.

Now, we will find the derivative of our given equation.

2x\cdot x'+x'y+xy'+4y\cdot y'=0

We have been given that y=2, x =2 and y'=-10.

2(2)\cdot x'+(2)x'+2(-10)+4(2)\cdot (-10)=0

4\cdot x'+2x'-20-80=0

6x'-100=0

6x'-100+100=0+100

6x'=100

\frac{6x'}{6}=\frac{100}{6}

x'=\frac{50}{3}

Therefore, the x-coordinate is increasing at a rate of \frac{50}{3} cm/sec.

7 0
3 years ago
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