Ask question

Ask question

All categories

3 years ago

12

Angel buys 6pumpkins for thanksgiving. He wants to make roasted pumpkin seeds. He realize that each pumpkin has about 756 seeds

inside of its cavity. Each member of his family will eat one pumpkin. About how many roasted pumpkin seeds will he have in all?
What do I need to find?

What information do I need to use?

What information do you NOT need to use?

How will I use the information?

1 answer:

Effectus [21]3 years ago

6 0

Answer:

There are pumpkin 756 seeds in all.

Step-by-step explanation:

Given:

Angel buys 6 pumpkins. Each pumpkin has about 756 seeds inside.

Now, to find total pumpkin seeds we multiply the total number of pumpkin from total number of seeds in each pumpkin has:

There are total 6 pumpkins.

In 1 pumpkin it has 756 seeds.

So, in 6 pumpkins there are = $756\times 6$

= $4536$ .

Therefore, there are pumpkin 756 seeds in all.

You might be interested in

A square with an area of 196 ft2 is reduced by a factor of 1/4. How long the new sides of the square?

soldier1979 [14.2K]

Answer:

$49$

Step-by-step explanation:

<h2><u>Lesson : Reducing by factors:</u></h2><h2><u /></h2>

Given :

$196ft^2$

$Reduced * factor * of * \frac{1}{4}$

When you have to reduce a number by a certain factor, you must divide the area with the amount that it is reducing from, therefore,

$\frac{196}{4}$

$49$

5 0

3 years ago

7.2x14.3 what is that?

Galina-37 [17]

Answer:

102.96

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

$T(1)=\frac{}{2+1}=\frac{1}{3}$

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

5 0

3 years ago

If point Q does not lie on the x-axis or the y-

Harman [31]

E. II or IV only because it will have different signs if point Q was in either of those quadrants

4 0

2 years ago

Which fraction is equivalent to 8

pentagon [3]

Answer:

16/2

Step-by-step explanation:There is a whole variety of fractions that simplify to 8, including this one. The trick is to find a numerator that is divisible by a denominator to get to 8.

3 0

3 years ago

Read 2 more answers