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3 years ago

7

Friday, Jason and Jeff went to the store to buy some bread. The total of $12 was divided equally among the friends. Jason paid $

4 and Jeff paid $4 as well. Who paid the other $4?

2 answers:

Marrrta [24]3 years ago

7 0

Answer:

their friend, friday

insens350 [35]3 years ago

6 0

Answer:

Friday

Step-by-step explanation:

it says friday, jason, and jeff

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ANSWER FOR BRAINLEST

ra1l [238]

Answer:

20π or 62.8

Step-by-step explanation:

Volume of a cone = 1/3πr²h

1/3(π)(2)²9

= 12π or 37.7

Volume of a hemisphere = 2πr²/3

2(π)(2)²

= 8π or 25.1

Total Volume = 20π or 62.8

6 0

3 years ago

Geometry question for homework.

mars1129 [50]

So remember that the distance formula is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .

1.

$\sqrt{(4-4)^2+(-2-(-5))^2}\\ \sqrt{0^2+3^2}\\ \sqrt{0+9}\\ \sqrt{9}\\ 3$

Distance between (4,-5) and (4,-2) is 3 units.

2.

$\sqrt{(1-(-1))^2+(1-4)^2}\\ \sqrt{2^2+(-3)^2}\\ \sqrt{4+9}\\ \sqrt{13}$

Distance between (1,1) and (-1,4) is √13 (or 3.61 rounded to the hundreths) units.

3 0

3 years ago

Simplify the product. (5 − 6)(2 + 7)

Nat2105 [25]

5 - 6 = -1
2 + 7 = 9
-1 × 9 = -9

-9 would be the simplified answer.

Hope this helps!

8 0

3 years ago

Help plz math question. EXPLAIN HOW YOU GOT THE ANSWER. 30 PTS btw thx if u help

pentagon [3]

Answer:

256

Step-by-step explanation:

4^4^4 = 4^(4^4) = 4^256

so written in base 4, there will be 256 zeroes after a 1.

This translates to approximately 154 digits in decimal.

Just like 2^2^2 = 2^4

there will be four zeroes after a 1.

2^2^2=2^(2^2) = 2^4 = 16 = 10000 (16 in base 2).

4 0

3 years ago

A Geometric series has a first term of 4 and a common ratio of 6. what formula is equivalent to the partial sum of the first n t

IRISSAK [1]

$\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=4\\
r=6
\end{cases}
\\\\\\
S_n=4\left( \cfrac{1-6^n}{1-6} \right)\implies S_n=4\left( \cfrac{1-6^n}{-5} \right)$

6 0

3 years ago