All categories

2 years ago

Write an equation that represents this comparison sentence. 24 is 4 times as many as 6

Mathematics

1 answer:

Nana76 [90]2 years ago

3 0

24 is 4 times as many as 6

6*4=24

You might be interested in

Design a logo for yourself or a fictional company using a variety of transformations. Describe each of the transformations you u

zavuch27 [327]

Answer:

Step-by-step explanation:

7 0

3 years ago

Michael played a game of darts. He threw 14 darts and hit the target

solniwko [45]

3/7

6 divided by 2 equals 3
14 divided by 2 equals 7

4 0

2 years ago

Question 6 of 10

Maksim231197 [3]

Answer:

(2.5 , 3.5)

Step-by-step explanation:

We can use the midpoint formula . Here the points are , (2,2) and (3,5) .

• Using Midpoint Formula :-

⇒ M = { (x1 + x2)/2 , (y1 + y2)/2 }

⇒ M = ( 2+3/2 , 5+2/2 )

⇒ M = ( 5/2 , 7/2 )

⇒ M = ( 2.5 , 3.5 )

<h3>Hence the midpoint is (2.5 , 3.5)</h3>

6 0

3 years ago

For working 13 hours you earn $58.50 and $121.50 for 27 hours. What is your rate of pay in dollars per hour?

Contact [7]

Y2-y1/x2-x1
121.5-58.5/27-13
63/14
$4.5 an hour.

8 0

3 years ago

Read 2 more answers

Someone help me out please

uysha [10]

1 step (B): raise both sides of the equation to the power of 2.

$(\sqrt{x+3}-\sqrt{2x-1})^2=(-2)^2,\\ (x+3)-2\sqrt{x+3}\cdot \sqrt{2x-1}+(2x-1)=4,\\ 3x+2-2\sqrt{x+3}\cdot \sqrt{2x-1}=4$ .

2 step (A): simplify to obtain the final radical term on one side of the equation.

$-2\sqrt{x+3}\cdot \sqrt{2x-1}=4-3x-2,\\ -2\sqrt{x+3}\cdot \sqrt{2x-1}=2-3x,\\ 2\sqrt{x+3}\cdot \sqrt{2x-1}=3x-2$ .

3 step (F): raise both sides of the equation to the power of 2 again.

$(2\sqrt{x+3}\cdot \sqrt{2x-1})^2=(3x-2)^2,\\ 4(x+3)(2x-1)=(3x-2)^2$ .

4 step (E): simplify to get a quadratic equation.

$4(2x^2-x+6x-3)=(3x)^2-2\cdot 3x\cdot 2+2^2,\\ 8x^2+20x-12=9x^2-12x+4,\\ x^2-32x+16=0$ .

5 step (D): use the quadratic formula to find the values of x.

$D=(-32)^2-4\cdot 16=1024-64=960, \\ \sqrt{D} =8\sqrt{5} ,\\ x_{1,2}=\dfrac{32\pm 8\sqrt{5}}{2} =16\pm 4\sqrt{5}$ .

6 step (C): apply the zero product rule.

$x^2-32x+16=(x-16-4\sqrt{5}) (x-16+4\sqrt{5}) ,\\ (x-16-4\sqrt{5}) (x-16+4\sqrt{5}) =0,\\ x_1=16+4\sqrt{5} ,x_2=16-4\sqrt{5}$ .

Additional 7 step: check these solutions, substituting into the initial equation.

3 0

3 years ago