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2 years ago

Anyone who answers gets 20 points.

Mathematics

1 answer:

goldenfox [79]2 years ago

5 0

Answer:answers what

Step-by-step explanation:

²³√∛×÷≈≠≤≥≡≅⇒,⇔∈∧∧∈∉∨∞αβ⇅ΔπФω↑↓∵∴↔→←⇵Δ⇅⇄⇆∫∑⊂⊃⊆⊇⊄⊅∀°∠∡⊥∪∩∅¬║⊕⊕⊕⊕∦∝∞∞∞㏒㏑

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What is 102% as a fraction

scZoUnD [109]

102% as a fraction would be 102/100, because a percent is by the one hundreds, and you can have a fraction be by hundreds. You can reduce the fraction by dividing it by 2, which equals 51/50. Another example, to clarify, is to make 41% a fraction, and decimal. Once again a percent is base 100, so the fraction would be 41/100, then to make it a decimal, you would put a decimal point in front of the 41, to make it .41. Hope I helped, and if you don't understand, tell me and I will try to make it better.

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3 years ago

Read 2 more answers

Hi help please! i’m struggling lol i already know the first part

sladkih [1.3K]

Answer:

180 Degrees minus 117= sixty three

answer is x is 63

Where did I get the 180?

- straight Angles have 180

- 117 is a part of 180

-so that is why you had to subtract

I hope this Helps You!!

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2 years ago

If 840 CLASS eighth graders, making up 4/5 of total eighth, studied algebra during summer, how many eighth graders are enrolled

Anarel [89]

Answer:

1008

Step-by-step explanation:

If 4/5 of eighth graders, or 80%, are enrolled with class, there are still 20% of students that didn't study. That means you add 168, the 20%, and you'll get 1008.

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2 years ago

Help me and I give you

Murljashka [212]

Im pretty sure its b im soo sorry if im wrong.

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3 years ago

Given that cot θ = 1/√5, what is the value of (sec²θ - cosec²θ)/(sec²θ + cosec²θ) ?

Bogdan [553]

Step-by-step explanation:

$\mathsf{Given :\;\dfrac{{sec}^2\theta - co{sec}^2\theta}{{sec}^2\theta + co{sec}^2\theta}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{{sec}\theta = \dfrac{1}{cos\theta}}}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{co{sec}\theta = \dfrac{1}{sin\theta}}}}$

$\mathsf{\implies \dfrac{\dfrac{1}{cos^2\theta} - \dfrac{1}{sin^2\theta}}{\dfrac{1}{cos^2\theta} + \dfrac{1}{sin^2\theta}}}$

$\mathsf{\implies \dfrac{\dfrac{sin^2\theta - cos^2\theta}{sin^2\theta.cos^2\theta}}{\dfrac{sin^2\theta + cos^2\theta}{sin^2\theta.cos^2\theta}}}$

$\mathsf{\implies \dfrac{sin^2\theta - cos^2\theta}{sin^2\theta + cos^2\theta}}$

Taking sin²θ common in both numerator & denominator, We get :

$\mathsf{\implies \dfrac{sin^2\theta\left(1 - \dfrac{cos^2\theta}{sin^2\theta}\right)}{sin^2\theta\left(1 + \dfrac{cos^2\theta}{sin^2\theta}\right)}}$