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2 years ago

Due in 10 min Need explination also first helpfull answer will be brainlyest

Mathematics

1 answer:

Doss [256]2 years ago

4 0

My answer is 25 so yeah

You might be interested in

If x\3-2=5x-2 then x=

nikdorinn [45]

X/3 -2 = 5x-2 (multiply by 3)

x-6=15x-6

x=15x

14x=0

x=0

6 0

2 years ago

How many factors does the following term have? 2xy

egoroff_w [7]

Answer:

4 factors.

Explanation:

Factors are numbers that if they are divided, then they would be divisible. We can see that the term is multiplied with 3 terms. Hence, they are definitely factors of that term. However, there is one more factor that is divisible by 2xy. The number 1 is divisible by 2xy. Hence, the factors of this term is 2, x, y, and 1.

Hoped this helped.

$BrainiacUser1357$

5 0

1 year ago

Can you please help me with this question please and i will give you a Brainiest right

kirill [66]

Answer:

135 grams

Step-by-step explanation:

you want to find the mass of only the grapes, not including the bowl. so in this case, you would subtract the mass of the bowl from the total mass. you do this because you dont need the mass of the bowl. 256g - 121g = 135g :)

4 0

2 years ago

Find all the complex roots. Write the answer in exponential form.

dezoksy [38]

We have to calculate the fourth roots of this complex number:

$z=9+9\sqrt[]{3}i$

We start by writing this number in exponential form:

$\begin{gathered} r=\sqrt[]{9^2+(9\sqrt[]{3})^2} \\ r=\sqrt[]{81+81\cdot3} \\ r=\sqrt[]{81+243} \\ r=\sqrt[]{324} \\ r=18 \end{gathered}$ $\theta=\arctan (\frac{9\sqrt[]{3}}{9})=\arctan (\sqrt[]{3})=\frac{\pi}{3}$

Then, the exponential form is:

$z=18e^{\frac{\pi}{3}i}$

The formula for the roots of a complex number can be written (in polar form) as:

$z^{\frac{1}{n}}=r^{\frac{1}{n}}\cdot\lbrack\cos (\frac{\theta+2\pi k}{n})+i\cdot\sin (\frac{\theta+2\pi k}{n})\rbrack\text{ for }k=0,1,\ldots,n-1$

Then, for a fourth root, we will have n = 4 and k = 0, 1, 2 and 3.

To simplify the calculations, we start by calculating the fourth root of r:

$r^{\frac{1}{4}}=18^{\frac{1}{4}}=\sqrt[4]{18}$

NOTE: It can not be simplified anymore, so we will leave it like this.

Then, we calculate the arguments of the trigonometric functions:

$\frac{\theta+2\pi k}{n}=\frac{\frac{\pi}{2}+2\pi k}{4}=\frac{\pi}{8}+\frac{\pi}{2}k=\pi(\frac{1}{8}+\frac{k}{2})$

We can now calculate for each value of k:

$\begin{gathered} k=0\colon \\ z_0=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{0}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{0}{2}))) \\ z_0=\sqrt[4]{18}\cdot(\cos (\frac{\pi}{8})+i\cdot\sin (\frac{\pi}{8}) \\ z_0=\sqrt[4]{18}\cdot e^{i\frac{\pi}{8}} \end{gathered}$ $\begin{gathered} k=1\colon \\ z_1=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{1}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{1}{2}))) \\ z_1=\sqrt[4]{18}\cdot(\cos (\frac{5\pi}{8})+i\cdot\sin (\frac{5\pi}{8})) \\ z_1=\sqrt[4]{18}e^{i\frac{5\pi}{8}} \end{gathered}$ $\begin{gathered} k=2\colon \\ z_2=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{2}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{2}{2}))) \\ z_2=\sqrt[4]{18}\cdot(\cos (\frac{9\pi}{8})+i\cdot\sin (\frac{9\pi}{8})) \\ z_2=\sqrt[4]{18}e^{i\frac{9\pi}{8}} \end{gathered}$ $\begin{gathered} k=3\colon \\ z_3=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{3}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{3}{2}))) \\ z_3=\sqrt[4]{18}\cdot(\cos (\frac{13\pi}{8})+i\cdot\sin (\frac{13\pi}{8})) \\ z_3=\sqrt[4]{18}e^{i\frac{13\pi}{8}} \end{gathered}$

Answer:

The four roots in exponential form are

z0 = 18^(1/4)*e^(i*π/8)

z1 = 18^(1/4)*e^(i*5π/8)

z2 = 18^(1/4)*e^(i*9π/8)

z3 = 18^(1/4)*e^(i*13π/8)

5 0

7 months ago

Suppose each of 12 players rolls a pair of dice 3 times. Find the probability that at least 4 of the players will roll doubles a

polet [3.4K]

Answer:

Our answer is 0.8172

Step-by-step explanation:

P(doubles on a single roll of pair of dice) =(6/36) =1/6

therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)

=1-(1-1/6)3 =91/216

for 12 players this follows binomial distribution with parameter n=12 and p=91/216

probability that at least 4 of the players will get “doubles” at least once =P(X>=4)

=1-(P(X<=3)

=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)

=1-0.1828

=0.8172

7 0

2 years ago