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3 years ago

PLEASE HELP ASAP GIVING 30 POINTS AND BRAINLIEST AND THANKYOU!!!!!!

Mathematics

2 answers:

trapecia [35]3 years ago

8 0

Part C. 19 and 8 are the 2 factors

WORK: since 56 + 96 = 152, and 152 goes into 8 19 times.

balu736 [363]3 years ago

7 0

Answer:

56 + 96 = 152 = 8 × 19

Step-by-step explanation:

The factors of 56 are:

1, 2, 4, 7, 8, 14, 28, 56

The factors of 96 are:

1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

The GCF of these 2 numbers is 8.

56 + 96 = 152

The factors of 152 are:

1, 2, 4, 8, 19, 38, 76, 152

We need to rewrite 152 as 2 factors, one of them is 8.

The other is the sum of 2 numbers that do not have a common factor.

The other factor is 19 because it is the sum of 10 and 9.

Factors of 10:

1, 2, 5, 10

Factors of 9:

1, 3, 9

Other than 1, they do not have any common factors.

152 = 8 × 19

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Order the group of parabolas from widest to narrowest.

Mamont248 [21]

Answer:

$\large\boxed{y=\dfrac{1}{4}x^2,\ y=-\dfrac{1}{2}x^2,\ y=\dfrac{3}{2}x^2}$

Step-by-step explanation:

The equation of a parabola: y = ax²

The larger the value of |a|, the narrower the parabola.

We have the following coefficients a:

$\left|\dfrac{1}{4}\right|=\dfrac{1}{4},\ \left|-\dfrac{1}{2}\right|=\dfrac{1}{2},\ \left|\dfrac{3}{2}\right|=\dfrac{3}{2},$

We arrange the coefficients from the smallest to the largest:

$\dfrac{1}{4}\$

Therefore you have the answer:

$y=\dfrac{1}{4}x^2,\ y=-\dfrac{1}{2}x^2,\ y=\dfrac{3}{2}x^2$

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3 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

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

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