All categories

3 years ago

How would I describe the steps it takes to solve this w+35-4=6

Mathematics

1 answer:

Butoxors [25]3 years ago

6 0

First step is combine your like terms:
w+31=6
Second step is subtract 31 from both sides:
w=-25

You might be interested in

Rewrite without parentheses and simplify (v-7)^2

Zina [86]

Your answer correct good night

8 0

2 years ago

Which graph comparing time and dollars would be best represented by an exponential function?

Rasek [7]

Answer:

x and y

Step-by-step explanation:

because it make sense

6 0

2 years ago

Read 2 more answers

Divided 2) Ellie made $253 for 22 hours of work. She makes $11 per hour so Identify the error:

patriot [66]

Answer:

Ellie should have made 253$ in 23 hours in time. If she worked for 22 hours, then she made 242$.

Step-by-step explanation:

253$ / 11$= 23 hours

22 hours X 11$= 242 hours

5 0

3 years ago

Help solving this problem, the second problem not the third

yawa3891 [41]

Given:

1 Van and 4 buses filled with 195 students.

1 Van and 5 buses filled with 241 students.

Let x and y be the number of students in Van and number of students in bus.

$\begin{gathered} x+4y=195\ldots\text{ (1)} \\ x+5y=241\ldots\text{ (2)} \end{gathered}$

Subtract equation (1) from equation (2)

$\begin{gathered} x+5y-x-4y=241-195 \\ y=46 \end{gathered}$

Substitute y=46 in equation (1)

$\begin{gathered} x+4(46)=195 \\ x+184=195 \\ x=195-184 \\ x=11 \end{gathered}$

Number of students that van can carry is 11.

Number of students that bus can carry is 46

7 0

1 year ago

Factor completely, then place the answer in the proper location on the grid.

Tresset [83]

$\mathrm{Factor\:out\:common\:term\:}9 \ \textgreater \ 9\left(x^4-25y^8\right)$

$Factor \ \textgreater \ x^4-25y^8$

$\mathrm{Apply\:difference\:of\:squares\:rule:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$
$x^2-5y^4=\left(x+\sqrt{5}y^2\right)\left(x-\sqrt{5}y^2\right)$

$x^2-5y^4 \ \textgreater \ \mathrm{Apply\:difference\:of\:squares\:rule:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$
$x^2-5y^4=\left(x+\sqrt{5}y^2\right)\left(x-\sqrt{5}y^2\right)$

$\left(x^2+5y^4\right)\left(x+\sqrt{5}y^2\right)\left(x-\sqrt{5}y^2\right)$

$9\left(x^2+5y^4\right)\left(x+\sqrt{5}y^2\right)\left(x-\sqrt{5}y^2\right)$

Hope this helps!

8 0

2 years ago