All categories

2 years ago

I need help on this math question

Mathematics

1 answer:

s2008m [1.1K]2 years ago

6 0

Answer:

A is correct. :)

Step-by-step explanation:

You might be interested in

How much $ do you make in an hour if you earn $42.25 in 6.5 hours?

mafiozo [28]

$6.50 per hour. 42.25/6.5=6.5

7 0

3 years ago

Can someone explain to me this derivative : y=3x√x

dimaraw [331]

Answer:

(9/2)√x.

Step-by-step explanation:

Convert the radical to an exponent.

x√x = x^1 * x^1/2

= x^(1 + 1/2)

= x^3/2

So the derivative of 3x^3/2 is found as follows:

y' = 3 * 3/2x^(3/2 - 1)

= (9/2)x^1/2

= (9/2)√x.

6 0

3 years ago

WILL GIVE BRAINIEST ANSWER!!! TUVW has vertices T(1,-3),U(3,-1),V(6,-4),and W(4,-6).Determine if quadrilateral TUVW is a paralle

Arte-miy333 [17]

Plotting the 4 points and doing visual inspection will tell us if the quadrilateral is a parallelogram. We can also do analytical techniques which are to determine the slope and the distance between the points. Doing either of these two will give us
<span>B. Yes,TUVW quadrilateral s a parallelogram because there are two pairs of parallel sides.
</span>

4 0

3 years ago

Ok. Last one please help me. I will give you 5 stars, and a like

____ [38]

Answer:

3.14

Step-by-step explanation:

The five dollars subtracting what he had left. so that gives you 3.14

☭

8 0

2 years ago

Read 2 more answers

A dealer sells a certain type of chair and a table for $40. He also sells the same sort of table and a desk for $83 or a chair a

vesna_86 [32]

$Chair=x$

$Table=y$

$Desk=z$

$\begin{Bmatrix}x+y&=&40\\y+z&=&83\\x+z&=&77\end{matrix}$

keep the first row as normal, then in the other ones, we can isolate Y and X

$\begin{Bmatrix}x+y&=&40\\y&=&83-z\\x&=&77-z\end{matrix}$

now we can replace at first row...

$\begin{Bmatrix}(77-z)+(83-z)&=&40\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}160-2z&=&40\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}-2z&=&40-160\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}-2z&=&-120\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}z&=&60\\y&=&83-z\\x&=&77-z\end{matrix}$

now we can replace the Z to discovery the other value

$\begin{Bmatrix}z&=&60\\y&=&83-60\\x&=&77-60\end{matrix}$

$\boxed{\boxed{\boxed{\begin{Bmatrix}Chair&=&17\\Table&=&23\\Desk&=&60\end{matrix}}}}$

6 0

3 years ago