Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
Triangle ABC is defined by the points A(3,8), B(7,5), and C(2,3).
Create an equation for a line passing through point A and perpendicular to BC.
y=
X +
Reset
Next
Answers
Answer:
y = - [tex]\frac{5}{2}[/tex] x + [tex]\frac{31}{2}[/tex]
Step-by-step explanation:
Calculate the slope m of BC using the slope formula
m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]
with (x₁, y₁ ) = B (7, 5 ) and (x₂, y₂ ) = C (2, 3 )
[tex]m_{BC}[/tex] = [tex]\frac{3-5}{2-7}[/tex] = [tex]\frac{-2}{-5}[/tex] = [tex]\frac{2}{5}[/tex]
Given a line with slope m then the slope of a line perpendicular to it is
[tex]m_{perpendicular}[/tex] = - [tex]\frac{1}{m}[/tex] = - [tex]\frac{1}{\frac{2}{5} }[/tex] = - [tex]\frac{5}{2}[/tex] , then
y = - [tex]\frac{5}{2}[/tex] x + c ← is the partial equation
To find c substitute A (3, 8 ) into the partial equation
8 = - [tex]\frac{15}{2}[/tex] + c ⇒ c = 8 + [tex]\frac{15}{2}[/tex] = [tex]\frac{16}{2}[/tex] +[tex]\frac{15}{2}[/tex] = [tex]\frac{31}{2}[/tex]
y = - [tex]\frac{5}{2}[/tex] x + [tex]\frac{31}{2}[/tex] ← equation of perpendicular line
The equation for a line passing through point A and perpendicular to BC is 2y = -5x +31 or y = -2.5x +15.5.
What is the equation of a Straight Line ?The equation of the straight line is represented by
y = mx+c
where m is the slope , and c is the intercept on y axis.
The equation of the line BC for points B(7,5), and C(2,3) is
y- y' = m(x-x')
m = 2/5
and c =
y-5 = (2/5)x - (14/5)
y = (2/5)x+(11/5)
The slope of the line perpendicular rot this line is given by
as the product of two perpendicular lines should be -1
m = -5/2
y = (-5/2)x +c
8 = (-15/2) + c
8 +(15/2) = 0
31/2 = c
Therefore the equation for a line passing through point A and perpendicular to BC is
2y = -5x +31
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36/18= 0.5
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Answer:
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Answer:
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Step-by-step explanation:
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What are Exponential Equations?An exponential equation is an equation with exponents where the exponent (or) a part of the exponent is a variable.
Standard exponential equation :
y = [tex]a(b)^{x} + k[/tex]
where a is the initial value
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k is constant
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now ,
Substituting the values
y = [tex]a(b)^{x} + k[/tex]
y = [tex]a(b)^{x} -5[/tex]
for point (0.-3)
-3 = [tex]a(b)^{0} -5[/tex]
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a = 2
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1 = [tex]2(b)^{1} -5[/tex]
1 = 2b - 5
b = 3
Therefore , equation of exponential graph is
y = [tex]2(3)^{x} -5[/tex]
Hence ,The equation for the graph is y = [tex]2(3)^{x} -5[/tex] .
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Step-by-step explanation:
I answered another similar question by you earlier, so I really hope you read through this explanation, as I will go far in depth. It's important you learn to do this so as not to simply skate through homework, then struggle with assessment.
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Answers
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Answer:
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[tex]-3\log_{3}\left(x-2\right)=-25[/tex]
Thanks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!...........
Answers
[tex] - 3 log_{3}(x - 2) = - 25 \\ = > log_{3}(x - 2) = \frac{ - 25}{ - 3} \\ = > log_{3}(x - 2) = \frac{25}{3} \\ = > (x - 2) = ( \frac{25}{3} ) ^{3} \\ = > x - 2 = \frac{15625}{27} \\ = > x = \frac{15625}{27} + 2 \\ = > x = \frac{15625 + 54}{27} \\ = > x = \frac{15679}{27} \\ = > x = 5226.33[/tex]
Answer:
5226.33 (approximate value)
Hope you could get an idea from here.
Doubt clarification - use comment section.