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GrogVix [38]
3 years ago
14

How do you find x and y

Mathematics
1 answer:
andriy [413]3 years ago
5 0
To find X and Y X is horizontal Y is vertical, so that means SCP is horizontal and EF or a B is oops sorry all the way around
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Simplify the following expression:
Morgarella [4.7K]

Answer:

2nd option

Step-by-step explanation:

Given

2x - 8y + 3x² + 7y - 12x ← collect like terms

= 3x² + (2x - 12x) + (- 8y + 7y)

= 3x² + (- 10x) + (- y)

= 3x² - 10x - y

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3 years ago
6. A study recorded the number of baskets scored each game by four basketball players. The MAD from this data is shown below. Wh
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D. Corey!

Step-by-step explanation:

I hope this helps! Have a great rest of your day!

5 0
3 years ago
Michael saves $423 dollars a month for college.
alexandr1967 [171]

Answer:

(a). $20,000

(b). The estimate will be lower than the actual amount.

Step-by-step explanation:

We have been given that Michael saves $423 dollars a month for college.

(a). We know that 1 year equals 12 months.

4 years = 4*12 months = 48 months.

\text{Actual amount saved}=\$423\times 48

Since we are asked to find the estimated amount of money Michael will save in 4 years, so we will estimate both quantities as:

423\approx 400\\\\48\approx 50

\text{Actual amount saved}=\$400\times 50

\text{Actual amount saved}=\$20,000

Therefore, Michael will save approximately $20,000 in 4 years.

(b).

The estimate will be lower than the actual amount as we rounded $423 down $23 to nearest hundred that is $400 and rounded 48 up 2 to nearest ten that is 40.

Therefore, the estimate will be lower than the actual amount.

5 0
3 years ago
Camden is going to invest $600 and leave it in an account for 12 years. Assuming the
UkoKoshka [18]

~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$920\\ P=\textit{original amount deposited}\dotfill & \$600\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &12 \end{cases}

920=600e^{\frac{r}{100}\cdot 12}\implies \cfrac{920}{600}=e^{\frac{3r}{25}}\implies \cfrac{23}{15}=e^{\frac{3r}{25}}\implies \log_e\left( \cfrac{23}{15} \right)=\log_e\left( e^{\frac{3r}{25}} \right) \\\\\\ \ln\left( \cfrac{23}{15} \right)=\cfrac{3r}{25}\implies 25\cdot \ln\left( \cfrac{23}{15} \right)=3r\implies \cfrac{25\cdot \ln\left( \frac{23}{15} \right)}{3}=r \\\\\\ 3.5620\approx r\implies \stackrel{\%}{3.56}\approx r

7 0
3 years ago
Please I need it urgent​
lisabon 2012 [21]

Answer:

Step-by-step explanation:

Hope this helps u !!

4 0
3 years ago
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