All categories

2 years ago

The table below shows two equations:

Mathematics

1 answer:

Nina [5.8K]2 years ago

6 0

Equation 1

Subtracting 5 from both sides we have

|2x - 3| = -1 - Since an absolute values must be positive this equation has no solutions.

Equation 2:-

|5x + 3| - 10 = 3

|5x + 3| = 13

either 5x + 3 = 13 or 5x + 3 = -13

so x = 2 or x = -16/5 = -3.2

The correct choice is B

You might be interested in

What is the advantage of using<br> scientific names, instead of common<br> names?

Aleksandr-060686 [28]

Answer:

the scientific name is the same name around the world and if you use common names they are different around the world

Step-by-step explanation:

8 0

3 years ago

Which expression is equivalent to five to the tenth power multiplied by five to the fifth

ioda

Answer:

30517578125

Step-by-step explanation:

do u mean multiplying it? if not then plz lmk so i could help

6 0

3 years ago

Read 2 more answers

Aubrey’s monthly bank statement shows a total of $51 in fees for ATM withdrawals. If Audrey made 17 withdraws, what is Audrey ch

Tpy6a [65]

Answer:

Audrey was charged $3 for each withdrawal.

Step-by-step explanation:

We have to calculate the amount of money charged on one transaction.

In order to find the amount for one ATM withdrawal, the total amount for ATM withdrawals will be divided by number of withdrawals.

Given

$Total\ ATM\ withdrawal\ fees = \$51\\Total\ withdrawals = 17\\Amount\ for\ one\ withdrawal = \frac{Total\ fees}{No.\ of\ withdrawals}\\= \frac{51}{17}\\= \$3$

Hence,

Audrey was charged $3 for each withdrawal.

7 0

3 years ago

Find the slope and y intercept of the line represented by each table

charle [14.2K]

Step-by-step explanation:

6 0

3 years ago

Please help me find limit

cluponka [151]

9514 1404 393

Answer:

-13/11

Step-by-step explanation:

Straightforward evaluation of the expression at x=1 gives (1 -1)/(1 -1) = 0/0, an indeterminate form. So, L'Hopital's rule applies. The ratio of derivatives is ...

$\displaystyle\lim_{x\to 1}\dfrac{n}{d}=\dfrac{n'}{d'}=\left.\dfrac{\dfrac{4}{3\sqrt[3]{4x-3}}-\dfrac{7}{2\sqrt{7x-6}}}{\dfrac{5}{2\sqrt{5x-4}}-\dfrac{2}{3\sqrt[3]{2x-1}}}\right|_{x=1}=\dfrac{4/3-7/2}{5/2-2/3}=\dfrac{8-21}{15-4}\\\\=\boxed{-\dfrac{13}{11}}$

4 0

3 years ago