HELP ME PLEASE!!!!!!!!!!!!!!!!!!!!!!!!

You have 6 pints of glaze. It takes 7/8 of a pint to glaze a bowl and 9/16 of a pint to glaze a plate.

son4ous [18]

Answer:

a. How many bowls could you glaze?

6 bowls

How many plates could you glaze?

10 plates

b. You want to glaze 5 bowls, and then use the rest for plates.

How many plates can you glaze?

2 plates

How much glaze will be left over?

8/9 of glaze would be left

c. How many of each object could you glaze so that there is no glaze left over? Explain how you found your answer.

4 4/23 bowls and 4 4/23 plates

Step-by-step explanation:

We are told in the question

Number of pints of glaze = 6 pints

Number of pints to glaze a bowl= 7/8 of pint

Number of pints to glaze a plate = 9/16 of a pint

a.

How many bowls could you glaze?

7/8 pint = 1 bowl

6 pints = x

Cross Multiply

x = 6 pints/ 7/8

= 6 × 8/7

= 48/7

= 6.8571428571 bowls

We can only glaze whole numbers of a bowl, hence, we can only glaze 6 bowls

How many plates could you glaze?

9/16 pint = 1 bowl

6 pints = y

y = 6 pints/ 9/16

= 6 × 16/9

= 10.666666667 plates

We can only glaze whole number of a plate, hence, we can only glaze 10 plates

b. You want to glaze 5 bowls, and then use the rest for plates.

1 bowl = 7/8 pints

5 bowls = z

Cross Multiply

z = 5 × 7/8pints

= 4 3/8 pints of glaze would be used for bowls

How many plates can you glaze?

The rest is for plates, hence:

The amount of glaze left for plates is calculated as:

6 pints - 4 3/8

1 5/8 pints of glaze would be left over to glaze plates.

So therefore,

9/16 pints = 1 plate

1 5/8 pints =

= 2 8/9

Hence we can only glaze 2 plates

How much glaze will be left over?

2 8/9 pints - 2 pints

= 8/9 pints of glaze.

c. How many of each object could you glaze so that there is no glaze left over?

We have 6 pints of glaze

Number of pints to glaze a bowl= 7/8 of pint

Number of pints to glaze a plate = 9/16 of a pint

Let the number of each objects be represented by x

7/8 × x + 9/16 × x = 6 pints

= 4 4/23 bowls and 4 4/23 plates

3 0

3 years ago

What is 8,663,943,526 rounded to the nearest million? A. 8,663,000,000 B. 8,663,944,000 C. 8,664,000,000 D. 9,000,000,000

maksim [4K]

The answer is
<span>C. 8,664,000,000</span>

8 0

3 years ago

Avery has 5 more than twice as many dolls as Caraleigh. Avery has 13 dolls in her collection. write an equation to represent the

tamaranim1 [39]

13=2c+5
-5 -5
8=2c
/2 /2
4=c
____
| |
| c=4 |
|____|

3 0

3 years ago

The bus will leave in 7 minutes. what time will it leave?

hjlf

Answer:

1:52

Step-by-step explanation:

The clock reads 1:45 so simply add 7 minutes to that

3 0

2 years ago

Read 2 more answers

"find the reduction formula for the integral" sin^n(18x)

dexar [7]

Let

$I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx$
For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
$\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$

For the remaining integral, we can integrate by parts, taking

$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
$\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)$

So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

7 0

3 years ago