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Kay [80]
3 years ago
8

1. Is the function below linear or non linear? y = -8X2 + 9??​

Mathematics
1 answer:
romanna [79]3 years ago
3 0
Linear because it’s in y=mx+b form
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Determine whether the given linear equations are parallel, perpendicular, or neither.
ladessa [460]

Answer:

A. Perpendicular

Step-by-step explanation:

Lines are perpendicular if their slopes are opposite reciprocals of each other. Opposite, meaning if positive, the other slope is negative, and if negative, the other slope is positive. Reciprocal meaning, the number is flipped upside down, turning fractions into whole numbers and vice versa.

1/9  

-1/9

-9

The slopes are perpendicular

4 0
3 years ago
What is the measure of x?
UkoKoshka [18]

Answer:

x=4

Step-by-step explanation:

(whole secant) x (external part) = (tangent)^2

(x+5) *x = 6^2

x^2 +5x = 36

Subtract 36 from each side

x^2 +5x - 36 = 0

Factor

( x-4) (x+9) = 0

Using the zero product property

x-4 = 0  x+9 =0

x = 4  x=-9

Cannot be negative since that is negative length

x=4

7 0
3 years ago
(x+3)(2x^2+9x-2) please help
Juliette [100K]

Answer:

23+152+25−6.

;))

4 0
3 years ago
Simplify 7 + (−3) <br> A -10<br> B -4<br> C 4<br> D 10
ANEK [815]

Answer:

4

Step-by-step explanation:

The plus and minus sign cancel each other out so it becomes 7-3=4

6 0
3 years ago
Read 2 more answers
Use lagrange multiplier techniques to find the local extreme values of f(x, y) = x2 − y2 − 2 subject to the constraint x2 + y2 =
dexar [7]
Given f(x,\ y)=x^2-y^2-2 subject to the constraint x^2+y^2=16

Let g(x,\ y)=x^2+y^2.

The gradient vectors of f and g are:

\nabla f(x,\ y)=\langle2x,-2y\rangle and \nabla g(x,\ y)=\langle2x,2y\rangle

By Lagrange's theorem, there is a number \lambda, such that

\langle2x,-2y\rangle=\lambda\langle2x,2y\rangle=\langle2\lambda x,2\lambda y\rangle

\lambda=\pm1

It can be seen that f(x,\ y)=x^2-y^2-2 has local extreme values at the given region.
6 0
3 years ago
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