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3 years ago

QUESTION 1 OF 5 10(23+ 7) = 100l + 100l What is the value of l in the equation above?

Mathematics

2 answers:

yawa3891 [41]3 years ago

6 0

Answer: 1.5

Step-by-step explanation:

(10×23) + (10×7)=200l

230+70=200l

300=200l

l=300 ÷ 200

cancel the zeros to make it easier

l=3÷ 2

l=1.5

aalyn [17]3 years ago

3 0

Answer:

1.5

Step-by-step explanation:

An equation 10(23+7)=100ℓ+100ℓ is given and we are asked to find the value of ℓ.

To solve the equation, we have to move all the values containing ℓ on one side and other values on other side of equality.

10(23+7)=100ℓ+100ℓ

Multiply 10 with 23 and 7 which gives

(10 * 23 ) + ( 10 * 7) = 200 ℓ

Solving the above equation,

230 + 70 = 200 ℓ

300 = 200 ℓ

=> ℓ = 300 / 200

=> ℓ = 3/2

=> ℓ = 1.5

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Which expression is a fourth root of -1+isqrt3?

aleksklad [387]

Answer:

Step-by-step explanation:

$\sf n^{th}$ roots of a complex number is given by DeMoivre's formula.

$\sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}$

Here, k lies between 0 and (n -1) ; n is the exponent.

$\sf -1 + i\sqrt{3}$

a = -1 and b = √3

$\sf \boxed{r=\sqrt{a^2+b^2}} \ and \ \boxed{\theta = Tan^{-1} \ \dfrac{b}{a}}$

$\sf r = \sqrt{(-1)^2 + 3^2}\\\\ = \sqrt{1+9}\\\\=\sqrt{10}$

$\sf \theta = tan^{-1} \ \dfrac{\sqrt{3}}{-1}\\\\ = tan^{-1} \ (-\sqrt{3})$

$\sf = \dfrac{-\pi }{3}$

n = 4

For k = 0,

$\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi }{12}+iSin \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]$