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lara31 [8.8K]
3 years ago
9

<Q and <R are complementary angles. If m<Q = (31 - 3x)° and m<R = (19x - 5)°, find m<R​

Mathematics
2 answers:
likoan [24]3 years ago
5 0

Answer:

Step-by-step explanation:

90=(31-3x)+(19x-5)

90=31-3x+19x-590=19x-3x+31-5

90=16x+26

90-26=17x

64=4x

x=4

m∠R=19(4)-5

m∠R=76-5

m∠R=71

Amanda [17]3 years ago
3 0

31-3x+19x-5=90

26+16x=90

-26 -26

--------------------

16x= 64

16x/16= 64/16

x=4

m <R=19 (4)-5= 71.

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Answer: when x = 7, y = 16

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When we have two sets of inversely related coordinates, x₁ and y₁,

and x₂ and y₂, we can use the product rule, shown below,

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<h2>x₁ y₁ = x₂ y₂</h2><h2 />

Here, we know that y = 14 when x = 8 so

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We want to know the value of y when x = 7.

So our other coordinates will be 7 and y.

So we have (8)(14) = (7)(y).

Simplifying, (8)(14) is equal to 112 and (7)(y) is equal to 7y.

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Now we simply solve for y by dividing both sides of the equation by 7.

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3 years ago
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iris [78.8K]

Answer:(-1,2)

Step-by-step explanation:

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3 years ago
What is the standard form of 2x2x2x2x2?​
cupoosta [38]

Answer:

The standard enthalpy of formation is the energy that would be involved if one mole of the substance were to be formed from its elements that were at standard conditions of temperature and pressure which are 25 degrees C and one atm pressure.

Step-by-step explanation:

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Read 2 more answers
Two consecutive even integers have a sum of 46
eduard
<span>Two consecutive even integers have a sum of 46 are</span> 22 and 24

hope it helps
5 0
3 years ago
Read 2 more answers
You decide to put $5000 in a savings account to save $6000 down payment on a new car. If the account has an interest rate of 7%
bezimeni [28]

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$6000\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}

\bf 6000=5000\left(1+\frac{0.07}{12}\right)^{12\cdot t}\implies \cfrac{6000}{5000}\approx (1.0058)^{12t}\implies \cfrac{6}{5}\approx(1.0058)^{12t} \\\\\\ \log\left( \cfrac{6}{5} \right)\approx \log[(1.0058)^{12t}]\implies \log\left( \cfrac{6}{5} \right)\approx 12t\log(1.0058) \\\\\\ \cfrac{\log\left( \frac{6}{5} \right)}{12\log(1.0058)}\approx t\implies 2.63\approx t\impliedby \textit{about 2 years, 7 months and 16 days}

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