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Butoxors [25]
4 years ago
12

Solve each equation -10=-14v+14v

Mathematics
1 answer:
Vinvika [58]4 years ago
3 0

Answer:

No solution.

Step-by-step explanation:

Simplify. Combine like terms:

-10 = -14v + 14v

-10 = (-14v + 14v)

-10 = (0)

-10 ≠ 0 ∴ no solution is your answer.

~

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zhuklara [117]
Assuming the possible digits are from 0-9, there are 10*9*8 possibilities which come out to 720
5 0
3 years ago
Read 2 more answers
Any tips on how to do this?? i have no clue how
Black_prince [1.1K]

Answer:

6.7

Step-by-step explanation:

For this questions we have the oppsotie side and need the adjacent

out of SOH, CAH, TOA we will sue TOA

tan(56)=10/x

tan(56)/10=1/x

10/tan(56)=x

x=6.745 which rounds to 6.7

8 0
3 years ago
An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What
galina1969 [7]

Answer:

Part A:

The probability that all of the balls selected are white:

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

P(D_i)=\frac{1}{6} for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}

P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}

P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}

P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}

P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}

P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0

Part A:

The probability that all of the balls selected are white:

P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find P(D_3|A)

The data required is calculated above:

P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

7 0
3 years ago
Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8​
Ivanshal [37]

It looks like the differential equation is

\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}

Factorize the right side by grouping.

xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)

xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)

Now we can separate variables as

\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx

Integrate both sides.

\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx

\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx

\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}

You could go on to solve for y explicitly as a function of x, but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0
2 years ago
Mr. Drysdale earned 906.25 in interest in one year on money that he had deposited in
VikaD [51]

Answer: Mr Drysdale deposited 14,500

Step-by-step explanation:

Using the simple Interest formula

Simple Interest = Principal(Initial Money) * Interest Rate * Time Period

I = P*R*T

I = PRT

To get P which is the initial money he deposited, we divide both sides by RT

I/RT = PRT/RT

I/RT = P

Therefore the money deposited

P = I/RT

Interest (I)= 906.25

Interest Rate (R)= 6.25% = { converting percentage to decimal} (6.25/100)= 0.0625

Time Period (T) = 1 year

Principal (P)= ?

P= I/RT

P= 906.25/(0.0625 * 1)

P= 906.25/0.0625

P= 1450

The money he deposited in his local bank is 14,500

5 0
4 years ago
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