A term may be written in index form to?

A couple quick algebra 1 questions for 50 points!

kkurt [141]

Answer:

Question 6) Neither (Not an inverse variation)

Question 7) Inverse variation

Question 8) Inverse variation

Step-by-step explanation:

These are the two formulas we need to know prior to solving the question.

Direct variation: c times x

Inverse variation: c times 1/x --> c/x

- Here, c is the constant

6. y = x + 7

This is neither. It is not in the form of a direct, or inverse variation, so it does not belong in any category.

7. 14 = xy

This equation is an inverse variation.

xy = 14

y = 14/x

14 is the constant, and the format is the same as for an inverse variation.

8. x = -8/y

This equation is an inverse variation.

yx = -8

y = -8/x

-8 is the constant, and the format is the same as for an inverse variation.

So we can conclude that questions 7 and 8 is an inverse variations, while question 6 isn't.

6 0

1 year ago

Read 2 more answers

F(x) = 3x - 2x - 5<br><br> Factoring quadratics

bagirrra123 [75]

Answer:

$(3x -5)(x+1)$

Step-by-step explanation:

Factoring $3x^{2} -2x-5=0$ results in $(3x -5)(x+1)$

4 0

1 year ago

What is the image of point (-7, 7) for mapping (x, y) , (x – 1, x + 6).

Ugo [173]

Because we have the map we just need to change the x with -7 and y with 7.
So the map for (-7, 7) -> (-7-1, 7+6)=(-8, 13)

The answer to your question is B. I hope that this is the answer that you were looking for and it has helped you.

5 0

2 years ago

Find the prod<br> (4x+5)(5x +9)<br> OA. 0.2

Sedaia [141]

It's simple.

25x²+61x+45

7 0

2 years ago

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

2 years ago