How do you solve for y in the equation: x/y = y

Use the information given to find a convenient class width. Then list the class boundaries that can be used to create a relative

notka56 [123]

Answer:

hello some information is missing below is the missing information

8 classes for n = 75 measurements; minimum value = 20; maximum value = 175

answer : width = 20

class boundary : 20,40,60,80,100,120,140,160,180

Step-by-step explanation:

Using the information given the convenient class width

= (maximum value - minimum value) / number of class

= ( 175 - 20 ) / 8

= 155 / 8 = 19.375 to the nearest multiple of 5 = 20

Using the minimum value as the smallest class boundary

the class boundary will be :

20,40,60,80,100,120,140,160,180

since the maximum value 175

6 0

3 years ago

1. convert 1011101 two to

max2010maxim [7]

Answer:

1a) 93ten

1b) 135eight

2a) 565ten

2b) 769ten

Step-by-step explanation:

1a)

1011101

7 digits long the first digit (the one that is most far right) is the ones digit or 2^0 digit and so the last digit (the most far left) is 2^6.

So to put the number in base 10 we do:

1(2^6)+0(2^5)+1(2^4)+1(2^3)+1(2^2)+0(2^1)+1(2^0)

=64+0+16+8+4+0+1

=64+16+8+4+1

=(64+16)+(8+4+1)

=80+13

=93

1b)

93

What is the hightest power of 8 in 93?

8^x=93

8^1=8

8^2=64

8^3 would be a factor of 8 too much

So how many 8^2 's are in 93?

64 goes into 93 how many times?

It's once with a remainder of 93-64=29.

So now 29...what is the highest power of 8 that goes into 29? 8^1=8.

How many times? 3(8)=24 with remainder 5.

So 93=1(8^2)+3(8^1)+5(8^0).

93=135eight

2a) 2341six

2(6^3)+3(6^2)+4(6^1)+1(6^0)

2(216)+3(36)+4(6)+1

432+108+24+1

540+25

565

2b) 30001four

3(4^4)+0(4^3)+0(4^2)+0(4^1)+1(4^0)

3(256)+0+0+0+1

768+1

769

5 0

3 years ago

The height of Maury’s room is 8.4 feet from the floor to the ceiling. Maury wants to install a ceiling fan that hangs 1.875 feet

deff fn [24]

Answer: yes it would

Step-by-step explanation: because 8.4-1.875 is 6.525 and Maury is 6.6

7 0

2 years ago

Read 2 more answers

3. How is the Fundamental Theorem of Algebra true for quadratic polynomials?

myrzilka [38]

Answer:

Step-by-step explanation:

Theorm-The Fundamental Theorem of Algebra: If P(x) is a polynomial of degree n ≥ 1, then P(x) = 0 has exactly n roots, including multiple and complex roots.

Let's verify that the Fundamental Theorem of Algebra holds for quadratic polynomials.

A quadratic polynomial is a second degree polynomial. According to the Fundamental Theorem of Algebra, the quadratic set = 0 has exactly two roots.

As we have seen, factoring a quadratic equation will result in one of three possible situations.

graph 1

The quadratic may have 2 distinct real roots. This graph crosses the

x-axis in two locations. These graphs may open upward or downward.

graph 2

It may appear that the quadratic has only one real root. But, it actually has one repeated root. This graph is tangent to the x-axis in one location (touching once).

graph 3

The quadratic may have two non-real complex roots called a conjugate pair. This graph will not cross or touch the x-axis, but it will have two roots.

6 0

3 years ago

Use the following function rule to find f(4). f(x) = 5 + 7 x

vladimir2022 [97]

Answer:

33

Step-by-step explanation:

put 4 in for x and then solve

5+7(4)

5+28

33

6 0

3 years ago

Read 2 more answers

How do you solve for y in the equation: x/y = y - 6