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2 years ago

PLEASE HELP PLEASE PLEASE HELP

Mathematics

2 answers:

antoniya [11.8K]2 years ago

5 0

Answer:

C. 5 boxes with 4 doughnuts left over

Step-by-step explanation:

A baker made 64 doughnuts at his bakery on Monday.

The baker wants to put his doughnuts into boxes. Each box holds 12 doughnuts. What is the total number of boxes the baker can fill, and the total number of doughnuts he will have left over?

Given that the baker has 64 doughnuts and each box holds 12 doughnuts, how many boxes can he fill and how many will he have left over?

To find this, we need to divide 64 by 12 using long division.

*View the attached picture*

Hope this helps!

madam [21]2 years ago

3 0

Answer:

c) 5 boxes 4 donuts remain.

pls give brainlyist

Step-by-step explanation:

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vaieri [72.5K]

Answer:

$\frac{2}{3}$

Step-by-step explanation:

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3 years ago

Stuart needs to run by the store on his way to school. He begins at home, which on a coordinate plane is represented by the poin

Elza [17]

-1,5 is the correct answer here. 0+2, 4-4 then 2-3 0+5.

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3 years ago

If (-6,-38) and (5,28) are two anchor points on the trend line, then find the equation of the line

Delicious77 [7]

Answer:

The equation of the line is y = 6x - 2 .

Step-by-step explanation:

Formula for the equation of line is.

$(y - y_{1}) = m (x - x_{1})$

Where m is the slope.

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

As given

If (-6,-38) and (5,28) are two anchor points on the trend line.

Than

$m = \frac{28 - (-38)}{5 - (-6)}$

$m = \frac{28 + 38}{5 + 6}$

$m = \frac{66}{11}$

m = 6

Put the value in the equation of a line.

$(y - (-38)) = 6 (x - (-6))$

$(y + 38) = 6 (x + 6)$

$(y + 38) = 6x + 6\times 6$

$(y + 38) = 6x + 36$

y = 6x +36 - 38

y = 6x - 2

Therefore the equation of the line is y = 6x - 2 .

7 0

3 years ago

I) Find the first 4 terms in the expansion of

Katyanochek1 [597]

Answer:

(i) $64+192 x^2+240 x^4+160x^6$

(ii) 1072

Step-by-step explanation:

Part (i)

Using Binomial series formula:

$(2+x^2)^6=2^6+6C1 \cdot 2^{6-1}\cdot x^2+6C2 \cdot 2^{6-2}\cdot(x^2)^2+6C3 \cdot 2^{6-3}\cdot(x^2)^3$

$=64+6 \cdot 32\cdot x^2+15 \cdot 16\cdot x^4+20 \cdot 8 \cdot x^6$

$=64+192 x^2+240 x^4+160x^6$

Part (ii)

$(1-\frac{3}{x^2})^2=(1-\frac{3}{x^2})(1-\frac{3}{x^2})$

$=1-\dfrac{6}{x^2}+\dfrac{9}{x^4}$

$(2+x^2)^6(1-\frac{3}{x^2})^2=(64+192 x^2+240 x^4+160x^6)\left(1-\dfrac{6}{x^2}+\dfrac{9}{x^4} \right)$

$=64-\dfrac{384}{x^2}+\dfrac{576}{x^4}+192x^2-1152+\dfrac{1728}{x^2}+240x^4-1440x^2+2160+...$

There is no need to keep expanding since the remaining will include the variable.

Therefore, the term independent of x = 64 - 1152 + 2160 = 1072

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3 years ago

What is the length of ab

Strike441 [17]

It is 56 times 343 yeah

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3 years ago