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2 years ago

160 bananas;20%decrease

Mathematics

2 answers:

Jobisdone [24]2 years ago

7 0

Answer:

128 banana's... I would work it all out for u but the teacher just came in my room and had a go

Natalka [10]2 years ago

6 0

Answer: 128

Step-by-step explanation:

Percent (%)' means 'out of one hundred':

p% = p 'out of one hundred',

p% is read p 'percent',

p% = p/100 = p ÷ 100.

20% = 20/100 = 20 ÷ 100 = 0.2.

100% = 100/100 = 100 ÷ 100 = 1.

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Heyy could you help me out with this problem???

vivado [14]

Answer:

Step-by-step explanation:

Because of all the variables in the side lengths, we don't have enough information to use side-angle-side similarity.

We do have two angles of one triangle congruent to two corresponding angles of the other triangle, so we can use angle-angle similarity.

Sides AE and AD are corresponding sides. Sides AC and AB are corresponding.

Answer: C

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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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3 years ago

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What is the mode of the data set?

Liula [17]

Answer:

mode is the number listed most often and data can have more then 1 mode

the numbers most listed in the image are: 5 and zero

Step-by-step explanation:

I hope this helped!

3 0

3 years ago

Read 2 more answers

What is the surface area of the triangular prism shown

Zigmanuir [339]

The surface area of the shape shown is 1008 square centimeters. I hope this helps!

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2 years ago

LAST ATTEMPT! MARKING AS BRAINLIEST!! ( simplifying radicals)

serious [3.7K]

$\huge \bf༆ Answer ༄$

Let's solve ~

$\qquad \looparrowright \sf - 7 \sqrt{80 {x}^{4} y {}^{2} }$

$\qquad \looparrowright \sf - 7 \sqrt{2 {}^{4} \times 5 \times {x}^{4} \times {y}^{2} }$

$\qquad \looparrowright \sf - 7 \times {2}^{2} \times {x}^{2} \times y \times \sqrt{5}$

$\qquad \looparrowright \sf - 28 {x}^{2} y \sqrt{5}$

Therefore, Option D is correct ~

8 0

2 years ago