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NemiM [27]
3 years ago
15

7 A rock climber that is 0.74 mile above sea level

Mathematics
2 answers:
Rina8888 [55]3 years ago
6 0
If I understand what kind of answer this question is asking for it should be .12 mile above sea level.
liq [111]3 years ago
6 0
Here is the equation.

.74+.18-.8=
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Identify whether the following
erik [133]
It opens downwards so it looks like this “n”. That is because a in the formula is a negative number in this situation
6 0
3 years ago
Which of the following is true of the discriminant for the graph below
aniked [119]

Answer:

Step-by-step explanation:

The discriminant is used to determine the number and nature of the zeros of a quadratic. If the discriminant is positive and a perfect square, there are 2 rational zeros; if the discriminant is positive and not a perfect square, there are 2 rational complex zeros; if the discriminant is 0, there is 1 rational root; if the discriminant is negative, there are no real roots.

The roots/solutions/zeros of a quadratic are where the graph goes through the x axis. Those are the real zeros, even if they don't fall exactly on a number like 1 or 2 or 3; they can fall on 1.32, 4.35, etc. They are still real. If the graph doesn't go through the x-axis at all, the zeros are imaginary because the discriminant was negative and you can't take the square root of a negative number. As you can see on our graph, the parabola never goes through the x-axis. Therefore, the zeros are imaginary because the discriminant was negative. Choice C.  Get familiar with your discriminants and the nature of quadratic solutions. Your life will be much easier!

4 0
3 years ago
7
Rus_ich [418]

Answer:

something.....................

7 0
3 years ago
Read 2 more answers
VEEL
Andre45 [30]

Answer:

a_n=-3(3)^{n-1} ; {-3,-9, -27,- 81, -243, ...}

a_n=-3(-3)^{n-1} ; {-3, 9,-27, 81, -243, ...}

a_n=3(\frac{1}{2})^{n-1} ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

a_n=243(\frac{1}{3})^{n-1} ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

a_n=-3(3)^{n-1}

At n=1,

a_1=-3(3)^{1-1}=-3

At n=2,

a_2=-3(3)^{2-1}=-9

At n=3,

a_3=-3(3)^{3-1}=-27

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

a_n=-3(-3)^{n-1}

At n=1,

a_1=-3(-3)^{1-1}=-3

At n=2,

a_2=-3(-3)^{2-1}=9

At n=3,

a_3=-3(-3)^{3-1}=-27

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

a_n=3(\frac{1}{2})^{n-1}

At n=1,

a_1=3(\frac{1}{2})^{1-1}=3

At n=2,

a_2=3(\frac{1}{2})^{2-1}=1.5

At n=3,

a_3=3(\frac{1}{2})^{3-1}=0.75

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

a_n=243(\frac{1}{3})^{n-1}

At n=1,

a_1=243(\frac{1}{3})^{1-1}=243

At n=2,

a_2=243(\frac{1}{3})^{2-1}=81

At n=3,

a_3=243(\frac{1}{3})^{3-1}=27

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0
3 years ago
Help I’ll mark you brainly!
Gnesinka [82]
Pls mark brainleist

7 0
3 years ago
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