3 years ago

Is this a function? {(-1,3); (-2,-3); (2,0); (0,2)}

Mathematics

1 answer:

Molodets [167]3 years ago

8 0

Yes, because the x’s aren’t repeated

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Which statement is not true?

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The true answer is A. all parallelograms are rectangles

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Jimmy gets a discount of 35% for a pair of sneakers with an original price of $70 he also gets a discount of 25% on a shirt with

Fed [463]

Answer:

$32.00

Step-by-step explanation:

$70 X .35 = $24.50 sneaker discount

$30 X .25 = $7.50 shirt discount

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3 years ago

Use the graphs of f(x) = 2

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Answer:

Step-by-step explanation:

The solution to f(x) = g(x) occurs where the two graphs cross. That's at the point (-2, -1).

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3 years ago

SHOW STEPS!! 4. x + 4y -- 67 = -1 2x - y + 2z = -7 -x + 2y -- 43 = 5

mafiozo [28]

1.x+4y- - 67 =-1
x+4y+67=-1
x+4y=-1-67
X+4y=-68

2.2x-y+2z=-7
2x=-7+y-2z
x=-7/2+1/2y-z
x=-7/2+1/2y-z, y€R, z€R

Warning: I don’t have the € and R

3.-x+2y- -43=5
-x+2y+43=5
-x+2y=5-43
-x+2y=-38
x-2y=38

Hope this helps!
Please mark me brainliest if possible

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3 years ago

Simplify (1-cos theta)(1+cos theta) over (1-sin theta) (1+ sin theta)

slega [8]

$\bf sin^2(\theta)+cos^2(\theta)=1\to 
\begin{cases}
cos^2(\theta)=1-sin^2(\theta)\\
sin^2(\theta)=1-cos^2(\theta)
\end{cases} \\\\
-------------------------------\\\\
\stackrel{\textit{difference of squares}}{\cfrac{[1-cos(\theta )][1+cos(\theta )]}{[1-sin(\theta )][1+sin(\theta )]}}\implies \cfrac{1^2-cos^2(\theta )}{1^2-sin^2(\theta )}\implies \cfrac{1-cos^2(\theta )}{1-sin^2(\theta )}
\\\\\\
\cfrac{sin^2(\theta )}{cos^2(\theta )}\implies tan^2(\theta )$

4 0

4 years ago