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suter [353]
3 years ago
6

CHECK ANSWERS EASY POINTS

Mathematics
2 answers:
olga55 [171]3 years ago
4 0

Answer:

Yes, you are correct.

Step-by-step explanation:

Nice penmanship by the way.

coldgirl [10]3 years ago
3 0
Yeah you right lol i thought u needed help
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Can someone check this for me?
Tresset [83]
Inside the square root shouldn't be negative, so x must be negative in order to form positive number inside the root.

if x = 0, y = 0
if y = -1, y = -1
if y = -4, y = -2
if y = -9, y = -3

the points lie on the graph (0,0), (-1,-1), (-4,-2), (-9,-3)
7 0
3 years ago
Graph g(x), where f(x)=2x-5 and g(x)=f(x+1)
tigry1 [53]

Answer:

Graph g(x) = 2x - 3

Step-by-step explanation:

Plug in (x+1) to f(x) = 2x - 5:

g(x) = 2(x+1) - 5 = 2x + 2 - 5

g(x) = 2x - 3

4 0
3 years ago
Read 2 more answers
Yolanda dug a hole in the ground that is deeper than -10 inches . She says the depth of the hole is greater than 10 inches. Is Y
denpristay [2]
No, she is not right
8 0
3 years ago
Consider the function f(x) = x^3 - 7x^2 - 2x + 14
mars1129 [50]

Answer:

There are 3 zeros, which are:

x=7,\:x=-\sqrt{2},\:x=\sqrt{2}

Step-by-step explanation:

Given the function

f\left(x\right)\:=\:x^3\:-\:7x^2\:-\:2x\:+\:14

To get the zeros of f(x), set y or f(x) =0

so

0=x^3-7x^2-2x+14

as

x^3\:-\:7x^2\:-\:2x\:+\:14=\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)

so

\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0

Using the zero factor principle:

\mathrm{ \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

so

x-7=0\quad \mathrm{or}\quad \:x+\sqrt{2}=0\quad \mathrm{or}\quad \:x-\sqrt{2}=0

solving

x-7=0

  • x=7

x+\sqrt{2}=0

  • x=-\sqrt{2}

x-\sqrt{2}=0

  • x=\sqrt{2}

Therefore, there are 3 zeros, which are:

x=7,\:x=-\sqrt{2},\:x=\sqrt{2}

8 0
3 years ago
Need help with my words in vocab​
Lerok [7]
Okey. But how and where
6 0
3 years ago
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