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3 years ago

Julianne desk is 75 centimeters long she say her desk is 7.5 meters long describe her error

Mathematics

2 answers:

allsm [11]3 years ago

8 0

First of all she said meters instead of centimeter.
and 2 she added a decimal point in between 7 and 5.

algol133 years ago

8 0

7.5 means it is 7 meters and 5 centimeters but 75cm is 0.75 or just 75cm

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The three oldest child need shoes.you find a store that has shoes on sale. All shoes in the store are a marked $10 per pair.you

bagirrra123 [75]

The best deal would be "three for the price of two" since buying one and 25 percent off would be 17.5 plus 10 for a third one is 27.5 compared to 20 for 3

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3 years ago

What is log15 2^3 rewritten using the power property?

OlgaM077 [116]

ANSWER

$log_{15}( {2}^{3} ) = 3 \: log_{15}( {2} )$

EXPLANATION

According to the power property of logarithms:

$log_{x}( {y}^{n} ) = n \: log_{x}( {y} )$

The given logarithm is

$log_{15}( {2}^{3} )$

When we apply the power property to this logarithm, we get,

$log_{15}( {2}^{3} ) = 3 \: log_{15}( {2} )$

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3 years ago

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In a bag of marbles you have 3 green, 2 yellow, 6 blue, and 9 red marbles. What is the probability of picking a yellow marble fr

slamgirl [31]

The answer is 2 in 20 or 1 in 10, all you have to do is add up the total number of marbles and then have the number of yellow in the total. So 2 in 20. Hope this helps you!

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

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The weight, w, of a spring in pounds is given by 0.9 times the square root of the energy, E, stored by the spring in joules. If

frutty [35]

For this case we have the following equation:
$w = 0.9* \sqrt{E}$
Where,
w: The weight of a spring in pounds
E: the energy stored by the spring in joules.
Substituting values we have:
$w = 0.9* \sqrt{12}$
Making the corresponding calculation:
$w=3.12$
Answer:
the approximate weight of the spring in pounds is:
$w=3.12$

4 0

3 years ago

Read 2 more answers