HELPP PLS ! THE ANSWER IS x=4 BUT ..

A person is to install five devices each 6 7/8 inches wide with 3 1/2 inches between switches. How much space will be needed to

Ilya [14]

D = 48.375 in or 48 3/8 in

Answer: 48 3/8 inches or 48.375 inches

4 0

3 years ago

Which of the following modifications to the list of assets and liabilities below would result in a positive net worth? Home owne

ratelena [41]

His home is worth 105,000. That increases his net worth (but the mortgage negates most of that).
His car is worth 26000 but the loan is worth 22000 which negates most of the value of the car.
Investment fund and savings increase his net worth, but the credit card eats up another 16000. Yes he can pay the credit card off monthly, but he should never have run it up as much as he has.

In actual fact, he has little of his own that could be called net worth.
The house is net 5000
The car is net 4000
The two small funds give 6000
So far the assets number 15000
But the credit card is a debt that wipes out everything else. His best course of action is pay off the card as quickly as possible. The interest is going to eat him alive.

8 0

3 years ago

Find DF D 20 E 10 F DF =

vichka [17]

Answer:

30

Step-by-step explanation:

it is right

8 0

3 years ago

Read 2 more answers

In a class of 30 students (x+10) study algebra, (10x+3) study statistics, 4 study both algebra and statistics. 2x study only alg

Vladimir [108]

Answer:

1. The Venn diagrams are attached

2. When the statistics students number = 10·x + 3, we have;

The number of students that study

a. Algebra = 128/11

b. Statistic = 213/11

When the statistics students number = 2·x + 3, we have;

The number of students that study

a. Algebra = 16

b. Statistic = 15

Step-by-step explanation:

The parameters given are;

Total number of students = 30

Number of students that study algebra n(A) = x + 10

Number of students that study statistics n(B) = 10·x + 3

Number of student that study both algebra and statistics n(A∩B) = 4

Number of student that study only algebra n(A\B) = 2·x

Number of students that study neither algebra or statistics n(A∪B)' = 3

Therefore;

The number of students that study either algebra or statistics = n(A∪B)

From set theory we have;

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

n(A∪B) = x + 10 + 10·x + 3 - 4 = 27

11·x+13 = 27 + 4 = 31

11·x = 18

x = 18/11

The number of students that study

a. Algebra

n(A) = 18/11 + 10 = 128/11

b. Statistic

n(B) = 213/11

Hence, we have;

n(A - B) = n(A) - n(A∩B) = 128/11 - 4 = 84/11

Similarly, we have;

n(B - A) = n(B) - n(A∩B) = 213/11 - 4 = 169/11

However, assuming n(B) = (2·x + 3), we have;

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

n(A∪B) = x + 10 + 2·x + 3 - 4 = 27

2·x+3 + x + 10= 27 + 4 = 31

3·x = 18

x = 6

Therefore, the number of students that study

a. Algebra

n(A) = 16

b. Statistics

n(B) = 15

Hence, we have;

n(A - B) = n(A) - n(A∩B) = 16 - 4 = 12

Similarly, we have;

n(B - A) = n(B) - n(A∩B) = 15 - 4 = 11

The Venn diagrams can be presented as follows;

6 0

3 years ago

Shane marched 3 m East and 6 m north. How far is he from his starting point. Round your answer to the nearest tenth.

Dennis_Churaev [7]

Answer:

The distance from his starting point is 6.7 m.

Step-by-step explanation:

Given:

Shane marched 3 m East and 6 m north.

Now, to get the distance from his starting point.

Let the distance from the starting point be $x.$

Now, to solve using pythagorean theorem:

$AB^2+BC^2=AC^2$

$6^2+3^2=x^2$

$36+9=x^2$

$45=x^2$

Using square root on both sides we get:

$6.71=x\\\\x=6.71\ m.$

So, the distance from the starting point nearest to tenth is 6.7 m.

Therefore, he is 6.7 m from his starting point.

6 0

3 years ago