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3 years ago

5

The price of a new cell phone is $199. Eduardo’s cell phone carrier is offering a 15% off coupon, and the store is offering a $5

0 rebate. How much will Eduardo pay for the cell phone?

1 answer:

Blizzard [7]3 years ago

4 0

The answer is $119.15

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In the year 2000 there were 8.7 deaths per 1000 residents in the United States is there were 281,421,906 residence in the US dur

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448370.58

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3 years ago

Find the value of x. Round the length to the nearest tenth.<br><br> ANSWER: A)7.2 ft

PSYCHO15rus [73]

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7.2

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2 years ago

What are vertical angles? What are their measurement?

Elodia [21]

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3 years ago

In the figure below, angle y and angle x form vertical angles. Angle y forms a straight line with the 60° angle and the 70° angl

Dmitry_Shevchenko [17]

<u>Answer:</u>

The value of x is 50°.

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

<u>We know that:</u>

x = y
2(70 + 60) + x + y = 360

<u>Work:</u>

2(130) + x + x = 360
=> 260 + 2x = 360
=> 2x = 360 - 260
=> 2x = 100
=> x = 50

Hence, <u>the value of x is 50°.</u>

Hoped this helped.

$BrainiacUser1357$

7 0

2 years ago

Read 2 more answers

How do you solve this question? Any help appreciated!

fiasKO [112]

$\bf \displaystyle \int\limits_{1}^{e}\cfrac{1}{t}\cdot dt\\\\
-------------------------------\\\\
\textit{doing substitution}\\\\
u=\cfrac{1}{t}\implies u=t^{-1}\implies \cfrac{du}{dt}=-t^{-2}\implies \cfrac{du}{dt}=-\cfrac{1}{t^2}\\\\\\ -t^2du=dt\\\\
-------------------------------\\\\
\displaystyle \int\limits_{1}^{e}u\cdot -t^2du\impliedby \textit{now, let's do some substitution on the "t"}\\\\
-------------------------------\\\\$

$\bf u=\cfrac{1}{t}\implies t=\cfrac{1}{u}\implies t^2=\cfrac{1^2}{u^2}\implies t^2=\cfrac{1}{u^2}\\\\
-------------------------------\\\\
\displaystyle \int\limits_{1}^{e}u\cdot -\cfrac{1}{u^2}\cdot du\implies -1\int\limits_{1}^{e}\cfrac{1}{u}\cdot du\implies \left. -ln|u| \cfrac{}{}\right]_1^e$

$\bf \left. -ln\left( \frac{1}{t} \right) \cfrac{}{}\right]_1^e\implies 
\left[ -ln\left( \frac{1}{e} \right) \right]-\left[ -ln\left( \frac{1}{1} \right) \right]\implies 
\left[ -ln\left( e^{-1}\right) \right]-\left[ -ln\left( 1\right) \right]
\\\\\\\
[-(-1)]-[-(0)]\implies 1-0\implies 1$

8 0

3 years ago