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3 years ago

What is the coefficient of y int the expression 2x4+3y

1 answer:

6 0

The coefficient of y = 3

coefficient = the constant in front of a variable
Constant means that the number will not change.

hope this helps

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51 is 33 1/3 of what number?

Makovka662 [10]

1700, just use the inverse operation. The problem would look like this ( not inverse operation ) x ÷ 33 1/3 = 51

All you have to do is multiply the two numbers.

51 x 33 1/3 = 1700

Now plug it back in and check

1700 ÷ 33 1/3 does indeed equal 51.

5 0

3 years ago

Hlp me plz i will give you guys 25 points

mart [117]

(A4+a2) hoped this helps

6 0

3 years ago

The square of a positive number is equal to its triple added to 28. What is this number?

sashaice [31]

hi brainly user! ૮₍ ˃ ⤙ ˂ ₎ა

⊱┈────────────────────────┈⊰

$\huge{ \rm{Answer:}}$

The roots of the equation are 7 and -4, because 7 - 4 = 3 and 7. - 4 = -28

Step-by-step explanation:

The positive number is equal to 7.

Assuming that the number is equal to "x" we have:

$\begin{gathered}x^2 = 3x + 28\\\\x^2 - 3x - 28 = 0\\\\a = 1\\\\b = -3\\\\c = -28\\\end{gathered}$

By the sum and product method, we have:

$\begin{gathered}S = -b / a\\\\S = -(-3) / 1\\\\S = 3\\\\\end{gathered}$

$\begin{gathered}P= c / a\\\\P = 28 / 1\\\\P = 28\\\\\end{gathered}$

4 0

2 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

In a candy mix there are 4 green bars for every 3 red bars. How many red bars are there if there are 200 green bars

erastovalidia [21]

Answer:

Since there are 4 green bars for every 3 red bars and we are trying to find the number of red bars if there are 200 green bars, we can create the ratio:

4 x : 3 y

Where x is equal to the number of green bars and y is the number of red bars.

We know the number of green bars is equal to 200, so we can divide it by 4, giving us:

200 /4=50

Then we can solve for y , the number of red bars.

// Multiple y by 50

3 ⋅ 50 = 150

So for every 4 green bars, there are 3 red bars.

For every 200 green bars, there are 150 red bars

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers