All categories

3 years ago

If a tablespoon held 1/16 how many would I need for 1 1/2 cups

Mathematics

2 answers:

MAVERICK [17]3 years ago

7 0

You would need 24 tablespoons.

$\huge{\tfrac{1.5}{\frac{1}{16}} = 24}$

Georgia [21]3 years ago

3 0

You will need 24 of the 1/16

You might be interested in

Which of the following is a monomial?

Dmitrij [34]

"mono" means 1 or single. In algebra, a monomial is any number that is alone or by itself. For this question, among the choices, the monomial is the number 4. Therefore the answer to this question is letter B. 4.

5 0

3 years ago

Yaro buys a baseball cap for $9.50. He also buys a new baseball. Yaro speads $13.50 altogether.

VMariaS [17]

Answer:

13.5 = 9.5 + x

x=4

Step-by-step explanation:

In this problem, the price of the baseball cap, which is $9.50, and a baseball, whose price is unknown (and therefore must be the variable), total $13.50. So, to set up the equation add the two prices and set them equal to the total, 9.5+x=13.5.

Then, using the subtraction property of equality subtract 9.5 from both sides. This means x=4, so the baseball cost 4 dollars.

4 0

3 years ago

Suppose A and B represent two different school populations where A > B and A and B must be greater than 0. Which of the follo

Ulleksa [173]

We can reject the last one: subtracting a non-zero value will result in a smaller value.

So now we have:

2(A + B)
(A + B)2
A2 + B2

If all of them are mulptiplications, then they are all equivalent!

I mean by this, if what you meant is this:


2*(A + B)
(A + B)*2
A*2 + B*2

If there is no sign, then the multiplication sign is implicit,

and all of these expressions say exactly the same: two of A and two of B.

6 0

3 years ago

2,17,82,257,626,1297 next one please ?

In-s [12.5K]

The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule $n^4+1$ . The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the n-th term in this sequence by $a_n$ , and denote the given sequence by $\{a_n\}_{n\ge1}$ .

Let $b_n$ denote the n-th term in the sequence of forward differences of $\{a_n\}$ , defined by

$b_n=a_{n+1}-a_n$

for n ≥ 1. That is, $\{b_n\}$ is the sequence with

$b_1=a_2-a_1=17-2=15$

$b_2=a_3-a_2=82-17=65$

$b_3=a_4-a_3=175$

$b_4=a_5-a_4=369$

$b_5=a_6-a_5=671$

and so on.

Next, let $c_n$ denote the n-th term of the differences of $\{b_n\}$ , i.e. for n ≥ 1,

$c_n=b_{n+1}-b_n$

so that

$c_1=b_2-b_1=65-15=50$

$c_2=110$

$c_3=194$

$c_4=302$

etc.

Again: let $d_n$ denote the n-th difference of $\{c_n\}$ :

$d_n=c_{n+1}-c_n$

$d_1=c_2-c_1=60$

$d_2=84$

$d_3=108$

etc.

One more time: let $e_n$ denote the n-th difference of $\{d_n\}$ :

$e_n=d_{n+1}-d_n$

$e_1=d_2-d_1=24$

$e_2=24$

etc.

The fact that these last differences are constant is a good sign that $e_n=24$ for all n ≥ 1. Assuming this, we would see that $\{d_n\}$ is an arithmetic sequence given recursively by