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kow [346]
3 years ago
13

QUICK!!!!!!!! PLSSS I NEED HELP!!!

Mathematics
2 answers:
coldgirl [10]3 years ago
4 0

Answer:

It's the 4th one

Step-by-step explanation:

yaroslaw [1]3 years ago
3 0
I hate think through math it’s the 4th one though.
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Use the graph of the derivative of f to locate the critical points x0 at which f has neither a local maximum nor a local minimum
jok3333 [9.3K]
<span>Critical points are where the derivative is 0, i.e. where it crosses the x - axis

The Critical points lies where the derivative is 0, while it crosses the x-axis, SO, in this case the choice 3 looks like best answer for this.
</span>
6 0
3 years ago
Find the quotient of 3 2/5 and 3 1/6 . Express your answer in simplest form.
ladessa [460]
Decimal Form: 1.07
Mixed Number:1 7/95


3 0
3 years ago
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Math help please thanks due soon
dezoksy [38]

Problem 11

Answer: Angle C and angle F

Explanation: Angle C and the 80 degree angle are vertical angles. Vertical angles are always congruent. Angle F is equal to angle C because they are alternate interior angles.

============================================

Problem 12

Answer: 100 degrees

Explanation: Solve the equation E+F = 180, where F = 80 found earlier above. You should get E = 100.

============================================

Problem 13

Answer: 80 degrees

Explanation: This was mentioned earlier in problem 11.

============================================

Problem 14

Answers: complement = 50, supplement = 140

Explanation: Complementary angles always add to 90. Supplementary angles always add to 180. An example of supplementary angles are angles E and F forming a straight line angle.

7 0
3 years ago
Given $a \equiv 1 \pmod{7}$, $b \equiv 2 \pmod{7}$, and $c \equiv 6 \pmod{7}$, what is the remainder when $a^{81} b^{91} c^{27}$
Blizzard [7]
\begin{cases}a\equiv1\pmod7\\b\equiv2\pmod7\\c\equiv6\pmod7\end{cases}

a^{81}\equiv1^{81}\equiv1\pmod7

b^{91}\equiv2^{91}\pmod7

2^{91}\equiv(2^3)^{30}\times2^1\equiv8^{30}\times2\pmod7
8\equiv1\pmod7
2\equiv2\pmod7
\implies2^{91}\equiv1^{30}\times2\equiv2\pmod7

c^{27}\equiv6^{27}\pmod7

6^{27}\equiv(-1)^{27}\equiv-1\equiv6\pmod7

\implies a^{81}b^{91}c^{27}\equiv1\times2\times6\equiv12\equiv5\pmod7
6 0
3 years ago
Answer this problem<br><br> x- 2 = -3x + 10
Semmy [17]
The correct answer is x = 3
6 0
2 years ago
Read 2 more answers
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