All categories

3 years ago

How do you solve this 21 = 14 + 7x

Mathematics

1 answer:

pav-90 [236]3 years ago

7 0

Answer:

subtract 14 from each side

7=7x

x=1

Step-by-step explanation:

plz brainliest!

You might be interested in

A rectangular family room is 18 feet long and 14 feet wide.

Firlakuza [10]

Answer:

4 pieces of chair rail are needed

Step-by-step explanation:

Step 1: Calculate the perimeter of the room

Perimeter of rectangle = 2L + 2B

Length = 18

Breadth = 14

Perimeter of room = 2(18) + 2(14)

Perimeter of room = 64 feet

Step 2: Calculate the pieces of chair rail needed

Perimeter of room = 64 feet

Length of each chair rail = 16 feet

Total chair rails needed = Perimeter of room/length of each chair rail

Total chair rails needed = 64/16

Total chair rails needed = 4

Therefore, Mr. Herzog needs 4 pieces of chair rails.

5 0

3 years ago

Suppose that an individual has a body fat percentage of 19.7% and weighs 157 pounds. How many pounds of her weight is made up of

dsp73

30.93 pounds is body fat.

5 0

4 years ago

Read 2 more answers

What is the answer to -(7c - 18) - 2c>0

IgorLugansk [536]

Answer:

The answer would be C < 2

4 0

3 years ago

1. Which of the following is true?

Charra [1.4K]

Answer:

Step-by-step explanation:

7 0

3 years ago

Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

4 0

2 years ago