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3 years ago

9

What is a common multiple for 8,12,10

2 answers:

masya89 [10]3 years ago

8 0

A common multiple is 2, because it fits into 8,12, and 10 evenly

vagabundo [1.1K]3 years ago

3 0

2 is a common multiple for 8,12,10

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How do you turn fractions into decimals?

Anvisha [2.4K]

Step 1: Find a number you can multiply by the bottom of the fraction to make it 10, or 100, or 1000, or any 1 followed by 0s.

Step 2: Multiply both top and bottom by that number.

Step 3. Then write down just the top number, putting the decimal point in the correct spot (one space from the right hand side for every zero in the bottom number)

7 0

3 years ago

the ratio of the number of adults to the number of children in a library is 12:5 if there are 45 children in the library how my

anygoal [31]

I’m guessing 108. 45 divided by 5 is 9, so 9 times 12 is 108.

5 0

3 years ago

Read 2 more answers

Use Newton's method with initial approximation x1 = 1 to find x2, the second approximation to the root of the equation x4 − x −

drek231 [11]

Answer:

$x_{2} = 0.0000$

Step-by-step explanation:

The formula for the Newton's method is:

$x_{i+1} = x_{i} + \frac{f(x_{i})}{f'(x_{i})}$

Where $f' (x_{i})$ is the first derivative of the function evaluated in $x_{i}$ .

$x_{i+1} = x_{i} + \frac{x_{i}^{4}-x_{i}-3}{4\cdot x_{i}^{3}-1}$

Lastly, the value of $x_{2}$ is determined by replacing $x_{1}$ with its numerical value:

$x_{2} = x_{1} + \frac{x_{1}^{4}-x_{1}-3}{4\cdot x_{1}^{3}-1}$

$x_{2} = 1.0000 + \frac{1.0000^{4}-1.0000-3}{4\cdot (1.0000)^{3}-1}$

$x_{2} = 0.0000$

8 0

3 years ago

Need ASAP if you don’t know the answer don’t answer

zubka84 [21]

Answer:

Hmm

Step-by-step explanation:

8 0

3 years ago

A new neighborhood activity complex is being built in erie. The perimeter of the rectangular playing field is 256 yards. The len

maw [93]

Answer:

The length of the rectangular playing field is 84 yards and the width is 44 yards

Step-by-step explanation:

Let

x ------> the length of the rectangular playing field

y -----> the width of the rectangular playing field

we know that

The perimeter of the rectangular playing field is equal to

$P=2(x+y)$

$P=256\ yd$

so

$256=2(x+y)$ ------> equation A

we have that

$x=2y-4$ -----> equation B

Solve the system by substitution

Substitute equation B in equation A and solve for y

$256=2(2y-4+y)$

$128=3y-4$

$3y=128+4$

$3y=132$

$y=44$

Find the value of x

$x=2(44)-4=84$

therefore

The length of the rectangular playing field is 84 yards and the width is 44 yards

4 0

3 years ago