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3 years ago

14

Emma says that Function A has a greater initial value. Is Emma correct? Justify your response

1 answer:

otez555 [7]3 years ago

3 0

Answer:

no function b is greater

Step-by-step explanation:

You might be interested in

Please help me with this!!<br> Thank u

topjm [15]

Answer:

Step-by-step explanation:

Left

When a square = a linear, always expand the squared expression.

x^2 - 2x + 1 = 3x - 5 Subtract 3x from both sides

x^2 - 2x - 3x + 1 = -5

x^2 - 5x +1 = - 5 Add 5 to both sides

x^2 - 5x + 1 + 5 = -5 + 5

x^2 - 5x + 6 = 0

This factors

(x - 2)(x - 3)

So one solution is x = 2 and the other is x = 3

Second from the Left

i = sqrt(-1)

i^2 = - 1

i^4 = (i^2)(i^2)

i^4 = - 1 * -1

i^4 = 1

16(i^4) - 8(i^2) + 4

16(1) - 8(-1) + 4

16 + 8 + 4

28

Second from the Right

This one is rather long. I'll get you the equations, you can solve for a and b. Maybe not as long as I think.

12 = 8a + b

<u>17 = 12a + b Subtract</u>

-5 = - 4a

a = - 5/-4 = 1.25

12 = 8*1.25 + b

12 = 10 + b

b = 12 - 10

b = 2

Now they want a + b

a + b = 1.25 + 2 = 3.25

Right

One of the ways to do this is to take out the common factor. You could also expand the square and remove the brackets of (2x - 2). Both will give you the same answer. I think expansion might be easier for you to understand, but the common factor method is shorter.

(2x - 2)^2 = 4x^2 - 8x + 4

4x^2 - 8x + 4 - 2x + 2

4x^2 - 10x + 6 The problem is factoring since neither of the first two equations work.

(2x - 2)(2x - 3) This is correct.

So the answer is D

5 0

2 years ago

Graph the solution of −5a+6>11.<br> Please, I need help!!!

horsena [70]

Answer:

$\large\boxed{a$

Step-by-step explanation:

$-5a+6>11\qquad\text{subtract 6 from both sides}\\\\-5a>5\qquad\text{change the signs}\\\\5a$

4 0

3 years ago

Find the product with the exponent in simplest form. Then, identify the values of x and y. 1 1 X 63.64 = 6 Y X = y =

Snezhnost [94]

Answer:

x = 7 y = 12

Step-by-step explanation:

4 0

2 years ago

What's the equation of the given graph?

kozerog [31]

Answer:

y=2x

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The area of ABED is 49 square units. Given AGequals9 units and ACequals10 units, what fraction of the area of ACIG is represent

Mrac [35]

Answer:

The fraction of the area of ACIG represented by the shaped region is 7/18

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

In the square ABED find the length side of the square

we know that

AB=BE=ED=AD

The area of s square is

$A=b^{2}$

where b is the length side of the square

we have

$A=49\ units^2$

substitute

$49=b^{2}$

$b=7\ units$

therefore

$AB=BE=ED=AD=7\ units$

step 2

Find the area of ACIG

The area of rectangle ACIG is equal to

$A=(AC)(AG)$

substitute the given values

$A=(9)(10)=90\ units^2$

step 3

Find the area of shaded rectangle DEHG

The area of rectangle DEHG is equal to

$A=(DE)(DG)$

we have $DE=7\ units$

$DG=AG-AD=9-7=2\ units$

substitute $A=(7)(2)=14\ units^2$

step 4

Find the area of shaded rectangle BCFE

The area of rectangle BCFE is equal to

$A=(EF)(CF)$

we have

$EF=AC-AB=10-7=3\ units$

$CF=BE=7\ units$

substitute

$A=(3)(7)=21\ units^2$

step 5

sum the shaded areas

$14+21=35\ units^2$

step 6

Divide the area of of the shaded region by the area of ACIG

$\frac{35}{90}$

Simplify

Divide by 5 both numerator and denominator

$\frac{7}{18}$

therefore

The fraction of the area of ACIG represented by the shaped region is 7/18

5 0

3 years ago