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3 years ago

Which expression is equivalent to |-5|+|3|? -8 -2 2 8

Mathematics

1 answer:

devlian [24]3 years ago

7 0

It would be 8 because |-5|+|3| is just 5+3

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What else is represented or means the same thing as a sum

My name is Ann [436]

Answer:

aggregate, total, totality, whole, summation, entirety, full, tally, all, body

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Can you please help me

vova2212 [387]

Answer:

i think it's 10

Step-by-step explanation:

i only had this lesson once ever soooo...

i tried tho

3 0

3 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

<u>Step-by-step explanation:</u>

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$

$\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}$

$\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark$

3 0

4 years ago

A standard deck of cards contains 52 cards. One card is selected from the deck. Compute the probability of randomly selecting a

lisabon 2012 [21]

P(selecting a 10)= 4/52
P(selecting a 2)= 4/52
P(selecting a 4)=4/52
P(selecting an Ace)=4/52
P(selecting a diamond)=13/52

3 0

3 years ago

The circumference of a circle is 11pi inches. What is the area, in square inches, of the circle? Express your answer in terms of

Eva8 [605]

$\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=11\pi \end{cases}\implies 11\pi =2\pi r\implies \cfrac{11\pi }{2\pi }=r\implies \boxed{\cfrac{11}{2}=r} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of a circle}\\\\ A=\pi r^2\qquad \qquad \implies A=\pi \left( \cfrac{11}{2}\right)^2\implies A=\pi \left( \cfrac{11^2}{2^2}\right)\implies A=\cfrac{121\pi }{4}$

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4 years ago