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2 years ago

The sum of two and three times the difference of a number and five is equal to four times the difference a number and one.

Mathematics

1 answer:

Andru [333]2 years ago

4 0

Answer:

$x=-9$

Step-by-step explanation:

We translate the phrase into an equation:

$2+3(x-5)=4(x-1)\\$

Then we simplify:

$2+3(x-5)=4(x-1)\\2+3x-15=4x-4\\-13+3x=4x-4\\-9=x$

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Pls help ASAP I will give brainleist answer

s2008m [1.1K]

Answer:

6 3/10 pounds

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

If the simple interest on $1000 for 2 years is $100, then what is the interest rate?

iogann1982 [59]

Answer:

R = 450%/year

Equation:

r = (1/t)(A/P - 1)

Converting r decimal to R a percentage

R = 4.5 * 100 = 450%/year

4 0

2 years ago

Read 2 more answers

An orchard has 651 orange trees. The number of rows exceeds the number of trees per row by 10. How many trees are there in a row

natima [27]

Answer:

The number of trees in each row is 21.

Step-by-step explanation:

Let us assume the number of trees in each row = m

So, the number of rows in the orchard = Number of trees in each row + 10

or, total number of trees in each row = (10 +m)

Now, Total number of Trees = Number of Rows x Number of trees in 1 row

⇒ 651 = m (m + 10)

$651 = m^2 + 10m\\\implies m^2 + 10m - 651 =0\\or, m^2 + 31m 21m - 651 =0\\or, m(m +31) -21(m+31) =0\\\implies (m +31) (m-21) = 0$

⇒ (m +31) = 0, or (m-21) = 0

⇒ m= -31, or m = 21

Now, as m = The number of trees in each row, so m ≠ -31

So, m = 21

Hence, the number of trees in each row is m = 21.

8 0

3 years ago

Show that in a group of five people (where any two people are either friends or enemies), there are not necessarily three mutual

sveta [45]

Answer:

Lets call the people A,B,C,D and E. We need to find an example where there are not 3 mutual friends and not three mutual enemies.

Lets start by assuming that A has 2 friends, B and C, and 2 enemies, D and E.

Since we want A,B and C not to be mutually friends, then B and C neccesarily have to be enemies.

Now, since B and C are enemies, they cant be enemies at the same time with D, and they also cant be enemies at the same time with E. Therefore one of them has to be friends with D and one has to be friends with E.

Also, since D and E are enemies with A, they neccesarily need to be friends.

From the fact that D and E are friends, we coclude that they cant have a friend in common, as a consequence, B has to be friends with one of D and E and C has to be friend with the other. We can assume that B is friend with D and C is friend with E. If we use that, we have the following friendship configuration

---------------------

Friends of A: B, C

Enemies of A: D,E

-------------------------

Friends of B: A,D

Enemies of B: C, E

------------------------

Friends of C: A,E

Enemies of C: B,D

------------------------

Friends of D: B,E

Enemies of D: A,C

----------------------------

Friends of E: C,D

Enemies of E: A,B

--------------------------

As you can check, there is not three mutual friends nor three mutual enemies.

8 0

3 years ago

When Emily arrives from school she spent a minute eating a snack then she spent 15 playing with her dog and 32 doing her chores

Neporo4naja [7]

The question is incomplete:

When Emily arrived home from school she spent a minutes eating a snack. Then, she spent 15 minutes playing with her dog and 32 doing her chores . Finally, Emily spent 45 minutes finishing her homework. How long had Emily been home when she finished?

Answer:

Emily had been home for 1 hour and 33 minutes.

Step-by-step explanation:

In order to find the answer, first you have to add up the time she spent in each activity to find the amount of minutes she had been at home when she finished:

1+15+32+45=93

Now, you can express the 93 minutes in hours considering that one hour has 60 minutes. According to this, Emily had been home for 1 hour and 33 minutes (93-60=33).

8 0

3 years ago