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salantis [7]
3 years ago
10

Use the graph of the quadratic function y = - 1x2 + x + 4 to

Mathematics
1 answer:
True [87]3 years ago
6 0

Answer: A, -2 and 4

Step-by-step explanation: cause it it you’re welcome

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About 18% of the population of a large country is nervous around strangers
sashaice [31]

Answer:

ok

Step-by-step explanation:

4 0
2 years ago
(a) What is wrong with the following equation? x2 + x − 42 x − 6 = x + 7 (x − 6)(x + 7) ≠ x2 + x − 42 The left-hand side is not
jeka94

Answer:

Attached please find answer

Step-by-step explanation:

= (x² + x - 42) / (x - 6)

= [x² + (7x - 6x) - 42] / (x - 6)

= [x² + 7x - 6x - 42] / (x - 6)

= [(x² + 7x) - (6x + 42)] / (x - 6)

= [x(x + 7) - 6(x + 7)] / (x - 6)

= [(x + 7)(x - 6)] / (x - 6)

= x + 7 → when x approches 6

= 6 + 7

= 13

Other method:

Lim   (x² + x - 42) / (x - 6)

x→ 6

When x → 6, there is an indeterminate expression because: 6 - 6 = 0

Do you know the l'Hôpital's rule, when x → a:

Lim [ f(x) / g(x) ] = Lim [ f'(x) / g'(x) ]

f(x) = x² + x - 42

f'(x) = 2x + 1

g(x) = x - 6

g'(x) = 1

= f'(x) / g'(x)

= (2x + 1)/1

= 2x + 1 → when x approches 6

= 12 + 1

= 13

3 0
2 years ago
Use the dot product to determine whether the vectors are parallel, orthogonal, or neither. v = (sqrt)2j, w = 4i
svetoff [14.1K]
Hello,

As \vec{i}*\vec{j}=0\\
\vec{v}*\vec{w}=\sqrt{2}*\vec{i}*4*\vec{j}=4*\sqrt{2}*0=0
Vectors are orthogonal.
4 0
3 years ago
Read 2 more answers
URGENT
ludmilkaskok [199]

Answer:

99.7%

Step-by-step explanation:

the standard deviation is 4 minuet but  there is always outliers so thats  why its not 100% of the time and the minutes are  coverd  by the deviation.  

6 0
3 years ago
Read 2 more answers
Find the missing side lengths. Leave your answers as radicals in simplest form.
yawa3891 [41]
C is your answer, you tell without solving
3 0
2 years ago
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