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hichkok12 [17]
3 years ago
7

What is the slope of a line perpendicular to a line with equation 6x-4y=30

Mathematics
1 answer:
Nana76 [90]3 years ago
6 0
I think the answer for 6x-4y=30 is -1.5
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suppose a varies jointly with b and c and inversely with d and a=400 when b=16, c=5 and d=2. write the equation that models the
Studentka2010 [4]
For this item, we represent first the proportionality constant by the variable k. Being directly proportional to both b and c, these variable will be in numerator and d will be in the denominator. That is,
                                  a = kbc / d
Substituting the known values,
                                 400 = k(16)(5) / (2) 
The value of k is 10. The equation that models the relationship is therefore,
                                   a = 10bc/d
4 0
3 years ago
a bag has 9 green marbles and 6 red marble. half of the red marbles are made of plastic. a marbles selected at random from the b
Bingel [31]
It will be 4.5 as its half.
5 0
3 years ago
Kala has scored 21,22,20,20 and 17 point in her five basketball game so far. How many points does she need to score in her next
Oliga [24]
Step 1: Add up all the scores.
Step 2: count the number of  scores.
step 3: divide the sum of the scores by the number of scores themselves.
step 4: The answer is 26 points.
8 0
3 years ago
What is the formula to find the length of an arc when given the central angle and radius?
vazorg [7]
s = \dfrac{n}{360^\circ}2 \pi r

where
s = arc length
n = measure of central angle
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3 0
3 years ago
The temperature function (in degrees Fahrenheit) in a three dimensional space is given by T(x, y, z) = 3x + 6y - 6z + 1. A bee i
madam [21]

You're looking for the extreme values of T(x,y,z)=3x+6y-6z+1 subject to x^2+y^2+z^2=9. The Lagrangian is

L(x,y,z,\lambda)=3x+6y-6z+1+\lambda(x^2+y^2+z^2-9)

with critical wherever the partial derivatives vanish:

L_x=3+2\lambda x=0\implies x=-\dfrac3{2\lambda}

L_y=6+2\lambda y=0\implies y=-\dfrac3\lambda

L_z=-6+2\lambda z=0\implies z=\dfrac3\lambda

L_\lambda=x^2+y^2+z^2-9=0

Substituting the first three solutions into the last equation gives

\dfrac9{4\lambda^2}+\dfrac9{\lambda^2}+\dfrac9{\lambda^2}=9

\implies\lambda=\pm\dfrac32

\implies x=1,y=2,z=-2\text{ or }x=-1,y=-2,z=2

At these points, we have

T(1,2,-2)=28

T(-1,-2,2)=-26

so the highest temperature the bee can experience is 28º F at the point (1, 2, -2), and the lowest is -26º F at the point (-1, -2, 2).

6 0
3 years ago
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