Which expression is equivalent to 1/4y−1/2? 1/4(y−1/2) 1/4(y−2) −1/4(y−12) −1/4(y−2)

Antonio uses a calculator to find 38 minus StartFraction 44 minus 16 over 4 EndFraction and gets a result of –10. Which statemen

frozen [14]

Answer:

The second statement is Correct.

Step-by-step explanation:

Consider the provided information.

It is given that Antonio uses a calculator to find $38-\frac{44-16}{4}$ and gets a result of –10.

$38-\frac{44-16}{4}=38-\frac{28}{4}$

$=38-7$

$=31$

So, the first statement is incorrect.

2nd statement: Antonio found the answer for $38-44-\frac{16}{4}$

$38-44-\frac{16}{4}=-6-4$

$38-44-\frac{16}{4}=-10$

The second statement is Correct.

Third statement: Antonio found the answer for $\frac{38-44-16}{4}$

$\frac{38-44-16}{4}=-5.5$

So, the third statement is incorrect.

Fourth statement: Antonio found the answer for $38+\frac{44-16}{4}$

$38+\frac{44-16}{4}=38+7$

$38+\frac{44-16}{4}=45$

So, the fourth statement is incorrect.

Hence, the second statement is Correct.

5 0

3 years ago

Read 2 more answers

He earns a fixed weekly salary of $300 plus a fee of $5 for each car he parks. His potential earnings for a week are shown in th

tamaranim1 [39]

I think what you are looking for is 5x=300 therefore x=60 so at 60cars he would start to make more.

3 0

2 years ago

Graph the hyperbola using the transverse axis, vertices, and co-vertices:

Reptile [31]

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

<u>Transverse axis</u>

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

a = 3 and b = 6

The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

<u>The vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

<u>The co-vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

And

a = 3 and b = 6

The co-vertices are

(h - a, k) and (h + a, k)

So, we have:

(0 - 3, 0) and (0 + 3, 0)

Evaluate

(-3,0) and (3, 0)

See attachment for the graph of the hyperbola