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3 years ago

5x+5=−4x−5 determine if solution

Mathematics

1 answer:

11111nata11111 [884]3 years ago

5 0

Answer:

x=-1.112

Step-by-step explanation:

Put all like terms on one side:

5x+4x=-5-5

Simplify:

9x=-10

Solve:

x=-1.112

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Write as an equation: 1/3 of a shipment of books weights 28 pounds.

FromTheMoon [43]

Answer:

Option D

Step-by-step explanation:

A third of a shipment of books is 28 pounds. <u>Assuming that 'f' will be the total weight of the entire shipment,</u> we would divide 'f' by 3 to get 28.

The answer should be D.

3 0

3 years ago

Richard's cumulative GPA for 3 semesters_ was 2.0 for 42 credits. His fourth semester GPA was 4.0 for 14 course units. What is R

Phoenix [80]

Answer:

Richard's cumulative GPA = 2.5

Step-by-step explanation:

Richard's cumulative GPA for 3 semesters was 2.0 for 42 credits.

Average = 2

Total credits = 42

Average = sum of terms/total no of terms

2 = sum of terms/42

sum of terms = 84 ....(1)

His fourth semester GPA was 4.0 for 14 course units

Average = 4

Total credits = 14

sum of terms = 56 .....(2)

Average for 4 semesters,

$A=\dfrac{84+56}{42+14}\\\\A=2.5$

Hence, Richard's cumulative GPA for all 4 semesters is 2.5.

8 0

3 years ago

Plz help ill give BRAINLIEST it for a test fast plz

Artemon [7]

Answer:

i believe it is C 7,24,25

Step-by-step explanation:

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6 0

4 years ago

Read 2 more answers

In a stand there are 15 books. 5 of them are Mathematicsbooks.

8090 [49]

Answer: At least 4/14 = 28.5%

Step-by-step explanation:

Select the location of a math book at the far end. The probability that the book next to it is a math book:

4 remaining math books, out of the 14 remaining books

In the middle, there are two potential spots adjacent to each book, so it could be higher - that’s why I say “at least” 4/14

4 0

3 years ago

Let {an} be a sequence that satisfies the recurrence relation an = an-1 + 3 for n = 1, 2, 3, ... , and suppose that a0 = 2. What

Arturiano [62]

Given:

$a_n=a_{n-1}+3$

$a_0=2$

To find:

The values $a_1,a_2,a_3$ .

Solution:

We have,

$a_n=a_{n-1}+3$

For $n=1$ ,

$a_1=a_{1-1}+3$

$a_1=a_{0}+3$

$a_1=2+3$

$a_1=5$

For $n=2$ ,

$a_2=a_{2-1}+3$

$a_2=a_{1}+3$

$a_2=5+3$

$a_2=8$

For $n=3$ ,

$a_3=a_{3-1}+3$

$a_3=a_{2}+3$

$a_3=8+3$

$a_3=11$

Therefore, $a_1=5,a_2=8,a_3=11$ .

3 0

3 years ago