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2 years ago

8

Eleni withdrew $45.00 from her savings account. she then used her debit card to buy groceries for $30.15. what was the total amo

unt Eleni took out her account?

2 answers:

evablogger [386]2 years ago

4 0

$45 + $30.15 = $75.15

SVEN [57.7K]2 years ago

3 0

$45.00+$30.15=$75.15

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Mademuasel [1]

Asking the Math Gods...

A=W*L

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Triangle JKL is transformed to create triangle J'K'L'. The angles in both triangles are shown.

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The correct statement is AngleL = 25° AngleL' = 25°.

We have given that,

Triangle JKL is transformed to create triangle J'K'L'.

We have to choose the correct option.

<h3>What is the information about the transformed triangle?</h3>

The following information should be considered:

In a rigid transformation, the image & pre-image are congruent.

Reflection, translation, and rotation are rigid transformations.

In a non-rigid transformation, the image and pre-image are similar.

Dilation is a non rigid transformation.

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If the corresponding sides are proportional, then it is a nonrigid transformation.

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1 year ago

given the side lengths of 6, 13, and 20, determine whether the triangle is acute , right , obtuse, or not a triangle.

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Not a triangle.

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2 years ago

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The simplest (and most commonly used) area calculations are for squares and rectangles. To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area

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3 years ago

Prove the trigonometric identity

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Man this one is a world of its own :D Just a quick question are you a fellow Add Math student in O levels i remember this question from back in the day :D Anyhow Lets get started

For this question we need to know the following identities:

$1+tan^{2}x=sec^2x\\\\1+cot^2x=cosec^2x\\\\sin^2x+cos^2x=1$

Lets solve the bottom most part first:

$1-\frac{1}{1-sec^2x} \\\\$

Take LCM

$1-\frac{1}{1-sec^2x} \\\\\frac{1-sec^2x-1}{1-sec^2x} \\\\\frac{-sec^2x}{1-sec^2x} \\\\\frac{-(1+tan^2x)}{-tan^2x}$

now break the LCM

$\frac{-1}{-tan^2x}+\frac{-tan^2x}{-tan^2x}\\\\\frac{1}{tan^2x}+1\\\\cot^2x+1$

because 1/tan = cot x

and furthermore,

$cot^2x+1\\cosec^2x$

now we solve the above part and replace the bottom most part that we solved with $cosec^2x$

$\frac{1}{1-\frac{1}{cosec^2x} } \\\\\frac{1}{1-sin^2x} \\\\\frac{1}{cos^2x}\\\\sec^2x$

Hence proved! :D

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2 years ago