Ask question

Ask question

All categories

3 years ago

8

How many groups of sixty are in two hundred forty-four

2 answers:

Naya [18.7K]3 years ago

7 0

4 groups of 60 in 244 because 60x4 is 240. There is 4 left over.

ss7ja [257]3 years ago

6 0

The answer to this question is 4 groups.

You might be interested in

0=4t-16t^2 Solve this please

vaieri [72.5K]

Answer:

$\large\boxed{t=0\ \vee\ t=\dfrac{1}{4}}$

Step-by-step explanation:

$4t-16t^2=0\qquad\text{divide both sides by 4}\\\\\dfrac{4t}{4}-\dfrac{16t^2}{4}=\dfrac{0}{4}\\\\t-4t^2=0\\\\t(1-4t)=0\iff t=0\ \vee\ 1-4t=0\\\\1-4t=0\qquad\text{subtract 1 from both sides}\\\\-4t=-1\qquad\text{divide both sides by (-4)}\\\\t=\dfrac{1}{4}$

7 0

3 years ago

5 Divided by 35.15?<br> Sorry if I spam a lot of questions it’s my final exam for school.

Orlov [11]

ANSWER: 0.142247510668563

5 0

3 years ago

Read 2 more answers

LII<br> Animals Seen in the Forest<br> 20<br> Number of Animals<br> Deer<br> Birds Bears Squirrels

Darya [45]

What’s the question?

4 0

4 years ago

2x + 1=4x-3 solve for x

loris [4]

Answer:

x = 2

Step-by-step explanation:

To start, you want to isolate x.

To do so, get all the x terms to one side.

2x + 1 = 4x - 3

Subtract -4x on both sides.

-2x + 1 = -3

Then, subtract 1 on both sides.

-2x = -4

Finally, divide -2 on both sides.

x = 2

3 0

3 years ago

What two rational expressions sum to 2x+3/x^2-5x+4

Anni [7]

Answer:

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}$

Step-by-step explanation:

Given the rational expression: $\frac{2x + 3}{x^2 - 5x + 4}$ , to express this in simplified form, we would need to apply the concept of partial fraction.

Step 1: factorise the denominator

$x^2 - 5x + 4$

$x^2 - 4x - x + 4$

$(x^2 - 4x) - (x + 4)$

$x(x - 4) - 1(x - 4)$

$(x- 1)(x - 4)$

Thus, we now have: $\frac{2x + 3}{(x- 1)(x - 4)}$

Step 2: Apply the concept of Partial Fraction

Let,

$\frac{2x + 3}{(x- 1)(x - 4)}$ = $\frac{A}{x- 1} + \frac{B}{x - 4}$

Multiply both sides by (x - 1)(x - 4)

$\frac{2x + 3}{(x- 1)(x - 4)} * (x - 1)(x - 4)$ = $(\frac{A}{x- 1} + \frac{B}{x - 4}) * (x - 1)(x - 4)$

$2x + 3 = A(x - 4) + B(x - 1)$

Step 3:

Substituting x = 4 in $2x + 3 = A(x - 4) + B(x - 1)$

$2(4) + 3 = A(4 - 4) + B(4 - 1)$

$8 + 3 = A(0) + B(3)$

$11 = 3B$

$\frac{11}{3} = B$

$B = \frac{11}{3}$

Substituting x = 1 in $2x + 3 = A(x - 4) + B(x - 1)$

$2(1) + 3 = A(1 - 4) + B(1 - 1)$

$2 + 3 = A(-3) + B(0)$

$5 = -3A$

$\frac{5}{-3} = \frac{-3A}{-3}$

$A = -\frac{5}{3}$

Step 4: Plug in the values of A and B into the original equation in step 2

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{A}{x- 1} + \frac{B}{x - 4}$

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}$

7 0

3 years ago