All categories

3 years ago

A farmer took 2/3 of the strawberries that he harvested to a market. At the market, the farmer sold 1/4

Mathematics

1 answer:

Aliun [14]3 years ago

5 0

Answer: If you are wondering how much he has left, there are 4/25 strawberries left

Step-by-step explanation:

You might be interested in

Hurry Please!

blagie [28]

I believe the answer should be 4 and 2

3 0

3 years ago

Please please help!!!!!

algol13

Answer:

Its 58

Step-by-step explanation:

Because the bigger triangle is a dilation of the smaller one. In other words only the sides get bigger, the angles stay the same.

5 0

3 years ago

Which decimal represents the fraction of cats in this group that are black?

Lubov Fominskaja [6]

Since there are 8 total, and 4 is the amount to find as a fraction, we simply divide 4 by 8 to get 4/8

3 0

3 years ago

O is the centre of the circle, EF is a tangent, angle BCE = 28°, angle ACD = 31°

andriy [413]

Answer

a. 28˚

b. 76˚

c. 104˚

d. 56˚

Step-by-step explanation

Given,

∠BCE=28° ∠ACD=31° & line AB=AC .

According To the Question,

a. the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.(Alternate Segment Theorem) Thus, ∠BAC=28°

b. We Know The Sum Of All Angles in a triangle is 180˚, 180°-∠CAB(28°)=152° and ΔABC is an isosceles triangle, So 152°/2=76˚

thus , ∠ABC=76° .

c. We know the Sum of all angles in a triangle is 180° and opposite angles in a cyclic quadrilateral(ABCD) add up to 180˚,

Thus, ∠ACD + ∠ACB = 31° + 76° ⇔ 107°

Now, ∠DCB + ∠DAB = 180°(Cyclic Quadrilateral opposite angle)

∠DAB = 180° - 107° ⇔ 73°

& We Know, ∠DAC+∠CAB=∠DAB ⇔ ∠DAC = 73° - 28° ⇔ 45°

Now, In Triangle ADC Sum of angles in a triangle is 180°

∠ADC = 180° - (31° + 45°) ⇔ 104˚

d. ∠COB = 28°×2 ⇔ 56˚ , because With the Same Arc(CB) The Angle at circumference are half of the angle at the centre

For Diagram, Please Find in Attachment

4 0

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

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

3 0

3 years ago

Read 2 more answers