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Svetradugi [14.3K]
3 years ago
14

Ten increased by the quotient of a number and 2 is -15

Mathematics
1 answer:
Vlad1618 [11]3 years ago
6 0
2x +10 = -15

2x = -25

2x/2 = -25/2

x = -25/2

hope this helps
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The city of Raleigh has 10500 registered voters. There are two candidates for city council in an upcoming election: Brown and Fe
Murrr4er [49]

Answer:

quantitative

Step-by-step explanation:

4 0
3 years ago
suppose you have a dime, two pennies, and a quarter. one of the pennies was minted in 1976, and the other one was minted in 1992
notsponge [240]

The reason that the answers in part (a) and part (b) are not the same is because Q1 and Q2 are treated as two different coins in part (a) whereas they are effectively treated as the same coin in part (b). The reason they are treated as one coin is because they both contribute the same amount to the sum of money.

<h3>How to Solve Counting Problems?</h3>

A) If you choose at least one coin, this means you could choose 1, 2, 3 or 4 coins. Let us label the dime as D, the penny as P, the quarter minted in 1976 as Q1 and the quarter minted in 1992 as Q2.

Now, if you choose one coin, you could choose either D, P, Q1, or Q2. This gives us 4 possible sets.

If you choose two coins, you choose the following sets of coins: DP, DQ1, DQ2, PQ1, PQ2, Q1Q2. This gives us 6 possible sets.

If you choose three coins, you could the following sets of coins: DPQ1, DPQ2, DQ1Q2, PQ1Q2. This gives us 4 possible sets.

If you choose four coins, you can only choose DPQ1Q2. This gives us 1 possible set.

Therefore, the total number of different sets of coins you can form is 4 + 6 + 4 + 1 = 15 different sets of coins can be formed.

b) If you choose at least one coin, this means you could choose 1, 2, 3 or 4 coins.

If you choose one coin you could choose either D, P, Q1, or Q2. However, since Q1 and Q2 give us the same sum, they are effectively the same set. This gives us 3 possible sums (ten cents, one cent, or twenty-five cents.)

If you choose two coins, you could choose the following sets of coins (since Q1 and Q2 are the same coin value, we can say Q for any instance where either one of these coins would be a possibility): DP, DQ, PQ, Q1Q2. This gives us 4 possible sums (11 cents, 35 cents, 26 cents, or fifty cents.)

If you choose three coins, you could choose the following sets of coins (since Q1 and Q2 are the same coin value, we can say Q for any instance where either one of these coins would be a possibility): DPQ, DQ1Q2, PQ1Q2. This gives us 3 possible sums (36 cents, 60 cents, or 51 cents).

If you choose four coins, you can only choose DPQ1Q2. This gives us 1 possible sum (61 cents.)

Therefore, the total number of different sums of coins you can form is 3 + 4 + 3 + 1 = 11 different sums of money can be produced.

c) The reason that the answers in part (a) and part (b) are not the same is because Q1 and Q2 are treated as two different coins in part (a) whereas they are effectively treated as the same coin in part (b). The reason they are treated as one coin is because they both contribute the same amount to the sum of money.

Read more about Counting Problems at; brainly.com/question/13875198

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3 0
1 year ago
Find the mass of the solid paraboloid Dequals=​{(r,thetaθ​,z): 0less than or equals≤zless than or equals≤8181minus−r2​, 0less th
Lubov Fominskaja [6]

Answer:

M = 5742π  

Step-by-step explanation:

Given:-

- Find the mass of a solid with the density ( ρ ):

                             ρ ( r, θ , z ) = 1 + z / 81

- The solid is bounded by the planes:

                             0 ≤ z ≤ 81 - r^2

                             0 ≤ r ≤ 9

Find:-

Find the mass of the solid paraboloid

Solution:-

- The mass (M) of any solid body is given by the following triple integral formulation:

                           M = \int \int \int {p ( r ,theta, z)} \, dV\\\\

- We can write the above expression in cylindrical coordinates:

                           M = \int\limits\int\limits_r\int\limits_z {r*p(r,theta,z)} \, dz.dr.dtheta \\\\M = \int\limits\int\limits_r\int\limits_z {r*[ 1 + \frac{z}{81}] } \, dz.dr.dtheta\\\\

- Perform integration:

                           M = \int\limits\int\limits_r{r*[ z + \frac{z^2}{162}] } \,|_0^8^1^-^r^2 dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{(81-r^2)^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{6561 -162r + r^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + 40.5 -r +\frac{r^2}{162} ] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{[ 121.5r-r^2 -\frac{161r^3}{162} ] } \, dr.dtheta\\\\

                           M = 2*\int\limits_0^\pi {[ 121.5r^2-r^3 -\frac{161r^4}{162} ] } |_0^6 \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 121.5(6)^2-(6)^3 -\frac{161(6)^4}{162} ] }  \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 4375-216 -1288] }  \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 2871] }  \, dtheta\\\\M = 5742\pi  kg              

- The mass evaluated is M = 5742π                      

8 0
3 years ago
Alyssa, Bart and Meghan are
adoni [48]

Hi,

Essentially, we are saying that Alyssa gets 5 parts of the whole thing (140,000), Bart gets 3 parts of the whole and Meghan gets 2 parts of the whole. The sum total of their parts = the whole. So now we need to find out how much one part is. Thus,"one part" should be the variable. Substitute p for "one part", so Alyssa gets 5p, Bart gets 3p, and Meghan gets 2p. Since the sum of the parts = the whole, your problem sets up as:  5p + 3p + 2p = 140,000  -->  10p = 140,000 --> 10p/10 = 140,000/10  -->  x = 14,000.

Alyssa gets 5p or (5 x 14,000)

Bart gets 3p or (3 x 14,000)

Meghan gets 2p or (2 x 14,000)

If the whole was a loss, it would be a negative, and instead of getting money, they would lose money.

3 0
2 years ago
Number 1 write in standard form correct answer will get brainliest
Fudgin [204]

Answer:

Step-by-step explanation:


5 0
3 years ago
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