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Aleks [24]
3 years ago
7

What is another point on the line?

Mathematics
1 answer:
Kamila [148]3 years ago
7 0

Answer:

B. (0,9)

Step-by-step explanation:

Start the point on (4,3), and becasue the slope is -3/2 you do rise over run. so you could go up 3 and left 2 or down 3 and right 2. i went up 3 and left to twice from point (4,3) and arrived at point (0,9). so the answer is B. (0,9)

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URGENT!! EASY IM DUMB MY LAST 2 QUESTION WILL FOREVER BE GRATEFUL PLS HELP WILL GIVE BRANLIEST!! AT LEAST TAKE A LOOK!!!! PLS I
BigorU [14]

Answer:

16. A

17. D

Step-by-step explanation:

16. By saying that a line intersects one plane and then another, you are saying that a line is existing on two planes. This is a direct contradiction to the statement.

17. The triangle is equilateral because syllogism is basically connecting the dots. If the angles in the triangle are all equal, it has all equal sides, and if it has all equal sides, then it is equilateral, therefore, it is D, not C.

4 0
3 years ago
Solve for all values of x by factoring. x^2-5x=-2x
-Dominant- [34]

Answer:

x = 0

x = 3

Step-by-step explanation:

x^2 - 5x = -2x\\x^2 - 5x + 2x = 0\\x^2 - 3x = 0\\x(x - 3) = 0\\x=0 , x-3=0\\x=0, x=3

4 0
3 years ago
What are the first five sequences to the equation f(1)=2/3,f(n)=f(n-1)x3/2
iren [92.7K]

Answer:

2/3,1,3/2,9/4,27/8

Step-by-step explanation:

f(1)=2/3

f(n)=f(n-1)×3/2

f(2)=f(2-1)×3/2=f(1)×3/2=2/3×3/2=1

f(3)=f(3-1)×3/2=f(2)×3/2=1×3/2=3/2

f(4)=f(4-1)×3/2=f(3)×3/2=3/2×3/2=9/4

f(5)=f(5-1)×3/2=f(4)×3/2=9/4×3/2=27/8

7 0
3 years ago
A particular isotope has a​ half-life of 74 days. If you start with 1 kilogram of this​ isotope, how much will remain after 150
shtirl [24]

\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}


\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}

6 0
3 years ago
Find the equation of the line specified. The slope is -7, and it passes through ( 5, -3).
Montano1993 [528]
The general equation of a line is
y= ax + b
where a is the slope
So a = -7
So y= -7x + b
But it passes through the point (5, -3)
Substitute the coordinates to calculate b:
So -3 = -7(5) + b
-3 = -35 + b
b= 35-3
b= 32
So the equation of the line of slope -7 and passing through the point (5, -3) is
y= -7x + 32
7 0
3 years ago
Read 2 more answers
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