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grin007 [14]
3 years ago
10

Could someone uh, help me? I’m a bit confused.

Mathematics
2 answers:
11111nata11111 [884]3 years ago
7 0
1
_
16 for the first one
allochka39001 [22]3 years ago
4 0
(a)= 1/6, 1/4, 1/3, and 1/2 I’m pretty sure.
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Which of the following
Ronch [10]

Answer:

Option A

Step-by-step explanation:

<u>Given equation is</u>

=> 3y = 6x + 3

<u>In slope-intercept form, it becomes</u>

=> 3y = 3(2x+1)

=> y = 2x+1

So, Slope = m = 2

<u><em>Parallel lines have equal slope, So any line parallel to the above line would have its slope equal to 2</em></u>

=> Line parallel to 3y = 6x + 3 is y = 2x + 10

3 0
3 years ago
Is 0.069 less than 0.07
Nuetrik [128]
Yes, 0.069 is less than 0.07~

First, we compare the "one" place: 0 = 0.

Second, we compare the "tenth" place: 0 = 0.

Next, we compare the "hundredth" place: 6 < 7.

Because 6 < 7, 0.069 < 0.07 as a result~
6 0
3 years ago
exercise 5 explain how you know the values of the digits in the number 58?then exercise 6 `explain why you should record a 1 in
melamori03 [73]
58 is 50 plus 8. So its 50 and 8.That is for the first one.
3 0
2 years ago
Gina puts $ 4500 into an account earning 7.5% interest compounded continuously. How long will it take for the amount in the acco
Elza [17]

~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$5150\\ P=\textit{original amount deposited}\dotfill & \$4500\\ r=rate\to 7.5\%\to \frac{7.5}{100}\dotfill &0.075\\ t=years \end{cases}

5150=4500e^{0.075\cdot t} \implies \cfrac{5150}{4500}=e^{0.075t}\implies \cfrac{103}{90}=e^{0.075t} \\\\\\ \log_e\left( \cfrac{103}{90} \right)=\log_e(e^{0.075t})\implies \log_e\left( \cfrac{103}{90} \right)=0.075t \\\\\\ \ln\left( \cfrac{103}{90} \right)=0.075t\implies \cfrac{\ln\left( \frac{103}{90} \right)}{0.075}=t\implies\stackrel{\textit{about 1 year and 291 days}}{ 1.8\approx t}

4 0
1 year ago
An observer took a time-exposure photograph of polaris and five nearby stars. how many hours were required to show their star pa
olga55 [171]
Photographing stars requires you to keep your exposure open for a long time. Taking pictures of star trails would take about 30 minutes to three hours. If you multiply that for the 6 stars stated above in the problem then you would be needing that much time if the stars were close and approximately 3 to 18 hours capture their star trails if they were far apart from each other.
5 0
2 years ago
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