If one (1) tablespoon equals three (3) teaspoons, how many tablespoons are in 13 teaspoons?

A box contains 8 two-inch screws. Four have a Phillips head and 4 have a slotted head. In how many ways can 4 screws be chosen s

hammer [34]

Answer:

There are $6\cdot 6=36$ different ways to choose 4 screws such that 2 have a Phillips head and 2 have a slotted head.

Step-by-step explanation:

If 4 screws must be chosen so that 2 have a Phillips head and 2 have a slotted head, then you have to choose 2 screws with a Phillips head from 4 screws with a Phillips head and 2 screws with a slotted head from 4 screws with a slotted head.

You can choose 2 screws with a Phillips head from 4 screws with a Phillips head in

$C^4_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!\cdot2!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot1\cdot2}=6$

different ways.

You can choose 2 screws with a slotted head from 4 screws with a slotted head in

$C^4_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!\cdot2!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot1\cdot2}=6$

different ways.

In total there are $6\cdot 6=36$ different ways to choose 4 screws such that 2 have a Phillips head and 2 have a slotted head.

7 0

3 years ago

PLZZZ HELP ASAP!!!!!!!

patriot [66]

Answer:

2. y = x + 16; 0.21y = 0.27x + 2.88

Step-by-step explanation:

You have two conditions:

Volume of 18 % + volume of 27 % = volume of 21 %
Mass of acid in 18 % + mass of acid in 27% = mass of acid in 21 %

For the first condition,

16 + x = y Transpose

(1) y = x + 16

=====

For the second condition

V₁c₁ + V₂c₂ = V₃c₃

16×0.18 + 0.27x = 0.21y

2.88 + 0.27x = 0.21y Transpose

(2) 0.21y = 0.27x + 2.88

The system of equations is y = x + 16; 0.21y = 0.27x + 2.88

4 0

3 years ago

A collection of quarters and dimes is worth $14.90. There are 80 coins in all. Find how many of each there are in the collection

sashaice [31]

The trick here is to relate the NUMBER of coins to each other in one equation, and then the VALUE of the coins in another equation. If I have 1 dime, that 1 dime is worth 10 cents. The number of dimes is obviously not equal to the value. Let's call quarters q and dimes d. The number of these 2 types of coins added together is 80 coins. So q + d = 80. Now, we know that quarters are worth .25 and dimes are worth .1, so we express a quarter's worth as .25q; we express a dime's worth as .1d. The value of the coins we have is 14.90. So that equation is .25q + .1d = 14.90. Let's solve the first equation for q. q = 80 - d. We can now use that as a substitution for q into the second equation, giving us an equation with only 1 unknown, d. .25(80-d) + .1d = 14.90. Distributing through the parenthesis we have 20 - .25d + .1d = 14.90. Combining like terms gives us - .15d = - 5.1. We will divide both sides by - .15 to get that the number of dimes is 34. If we had a total of 80 coins, then the number of quarters is 80 - 34, which is 46. 46 quarters and 34 dimes

7 0

3 years ago

Raul raised 4 to a power, multiplied the result by 5 and then added 1. The result was 321. to what power did raul raise 4?

Mashutka [201]

P is the power.

(4^p)•5+1=321

Solve for p.