V=1 ; v=1-11) Zv+2v 3v -40 = A)5 B)-2 C)-8 D)3

Which of the following will result in a rational number? Select all that apply.

Burka [1]

Answer:

A

Step-by-step explanation:

the answer is a :

6 4/5+√9/4

8 0

2 years ago

Marissa dug three fifths of a hole in one tenth of an hour. Find the rate in holes per hour

snow_lady [41]

Answer:

Step-by-step explanation:

This might be easier to do in decimals. Let's try it.

3/5 = 0.6

1/10 = 0.1

rate = holes / hours

rate = 0.6/0.1

rate = 6 holes per hour

Try it with fractions.

rate = $\frac{\frac{3}{5} }{\frac{1}{10} }$ Invert and multiply the 1/10

rate = $\frac{3}{5} *\frac{10}{1}$ = 30/5 = 6

It's the same answer. Pick the method that you feel most comfortable with.

3 0

3 years ago

Nsider the following system of equations y=6x^2+1, y=x^2+4

sp2606 [1]

The given equations are y= 6x^2+1 and y=x^2+4

Both the equations equal y so we can make them equal

6x^2+1=x^2+4

To bring x term on one side we subtract x^2 both sides

6x^2-x^2+1=4

5x^2+1=4

To isolate x term we subtract 1 both sides

5x^2=4-1

5x^2=3

Dividing by 5 both sides

X^2=3/5

Taking root of both sides we have:

X=±√⅗

Substituting x value to find y

When x=+√⅗=0.7745

Y= 6x^2+1

Y=6(3/5)+1

Y= 18/5+1

Y=23/5 =4.6

When x= -√⅗=-0.7745

Y= 6(3/5)+1

Y=23/5=4.6

Point of intersections are ( 0.7745,4.6) and (-0.7745,4.6)

7 0

3 years ago

Pls help or explain steps

Tems11 [23]

Solving for a variable means to rewrite the equation so that the desired variable is alone on one side. So, we have to "move" all the other variables/constants in order to make this happen. We are allowed to add/subtract the same quantity from both sides, or to multiply/divide both sides by any non-zero number.

Let's start with the first example: we are given

11. $d=rt$

the variable $r$ is not alone on the right hand side because it is multiplied by $t$ . We can reverse this by dividing both sides by $t$ : the equation becomes

$\dfrac{d}{t}=\dfrac{rt}{t}=r$

So, we have written the equation as $r=\frac{d}{t}$ , which means that $r$ now sits alone on one side of the equation, and thus we have solved the equation for $r$ !