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avanturin [10]
3 years ago
11

PLEASE HELP ASAP NEED TO GET GRADE UP AND I NEED ANSWERS!!!!

Mathematics
2 answers:
skelet666 [1.2K]3 years ago
8 0

Answer:

1. B) 7

2. B) {y|y ≤ -1}

Step-by-step explanation:

1. Put -4 where x is in the function definition and do the arithmetic.

... 3 - (-4) = 3 +4 = 7

2. If the maximum function value is -1 and the values extend to -∞ from there, then the range is (-∞, -1]. In your notation, that is ...

... {y | y ≤ -1}

marishachu [46]3 years ago
8 0

B and B

f(x) → 3 - x

to evaluate f(- 4 ) substitute x = - 4 into f(x)

f(- 4 ) = 3 - (- 4 ) = 3 + 4 = 7 → B

the range of a function are the values of y for the function

this function ( probably quadratic ) has a vertex at (0, - 1), that is the y-value is - 1

Since the function opens down then the values of y are tending to negative infinity

so the range of values for y are less than or equal to - 1 to negative infinity

{y | y ≤ - 1 } → B


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Juliette ate 1/3 of a cake, and Brad ate 3/8 of the same cake. What fraction of the cake was eaten?
solniwko [45]

Answer:

17/24

Step-by-step explanation:

Find common denominators...

1/3 = 8/24

3/8 = 9/24

Now add...

17/24 of the cake was eaten

Hope this helps :)

5 0
2 years ago
Find all the zeros of the equation x^4-6x^2-7x-6=0
rusak2 [61]

Answer:

The zeros are

x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}

Step-by-step explanation:

We have been given the equation x^4-6x^2-7x-6=0

Use rational root theorem, we have

a_0=6,\:\quad a_n=1

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:3,\:6,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:3,\:6}{1}

-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2

=\left(x+2\right)\frac{x^4-6x^2-7x-6}{x+2}\\

=x^3-2x^2-2x-3

Again factor using the rational root test, we get

=\left(x+2\right)\left(x-3\right)\left(x^2+x+1\right)

Using the zero product rule, we have

x+2=0:\quad x=-2\\x-3=0:\quad x=3\\x^2+x+1=0\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\\\=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}

Therefore, the zeros are

x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}


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3 years ago
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Answer:

c

Step-by-step explanation:

Subtract 100 from both sides to get x by itself.

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