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olga55 [171]
2 years ago
15

A line goes through the points (8,11) and (-2,7). What is the slope of the line? Show your work. Write the equation of the line

in point-slope form. Show your work. Write the equation of the line in slope-intercept form. Show your work.
Mathematics
1 answer:
Iteru [2.4K]2 years ago
8 0

Answer:

Step-by-step explanation:

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What is 0.24% of 77? Round to the nearest hundredth
Kruka [31]

Answer:

.18

Step-by-step explanation:

77 x .24% (0.0024) = .1848

Round to the nearest hundredth...

.18!

6 0
3 years ago
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An extremely handsome math teacher was on a farm and he counted 32 animals; all pigs and chickens. Then he counted the number of
statuscvo [17]

Answer:

12 pigs and 20 chickens

Step-by-step explanation:

Chicken = 2 legs

Pig = 4 legs

(15 * 4) + (17 * 2) = 88                          NO

(12 * 4) + (20 * 2) = 88                         Yes, and 12 + 20 = 32

7 0
2 years ago
Divide 4/5 by 2/3.<br> a. 8/15b. 5/6c. 1 1/5d. 1 7/8
Ivan
4/5÷2/3=x
4/5×3/2=x
   12/10=x 
    1 1/5=x
5 0
3 years ago
A rectangular prism has a volume of 16cm^3 and a total surface area of 40cm^2. A similar prism has length 10 in., width 5 in. an
krok68 [10]
The volume is twice that of the cube of width, so the width of the original prism is
.. ∛(16 cm³/2) = 2 cm

Length = 4 cm
Width = 2 cm
Height = 2 cm
5 0
3 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
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