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ivann1987 [24]
3 years ago
14

When is g(x) = 0 for the function g(x) = 5⋅2^3x + 4?

Mathematics
1 answer:
SashulF [63]3 years ago
6 0
The answer should be when X=1
Hope this helped u out if not sorry
:D

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The maximum height reached by the barnacle is _______ m.
Oksanka [162]

Answer:

1

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Karl is making a pot of chili. The recipe calls for StartFraction 3 Over 8 EndFraction cup of chili powder, but Karl only wants
mina [271]

Given:

The recipe calls for \dfrac{3}{8} cup of chili powder.

Karl only wants to use half as much so it won’t be so spicy.

To find:

How much chili powder should Karl use?

Solution:

We have,

Chill power for recipe = \dfrac{3}{8} cup

Chill power used by Karl is half  of chill power for recipe.

\text{Chill power used by Karl}=\dfrac{1}{2}\times \dfrac{3}{8} cup\text{Chill power used by Karl}=\dfrac{3}{16} cup

Therefore, \dfrac{3}{16} cup of chili powder should Karl use.

7 0
2 years ago
Maths question I'm struggling with.
Alex17521 [72]

Answer:

A = 2.63

B = 2

Step-by-step explanation:

To find the population density, what we need to do simply is divide the population by the area of the UK

mathematically, that would be;

(6.41 * 10^7)/243,610 = 263.125

To the nearest whole person= 263 people/ square meter

So now we want to write 263 in standard form = 2.63 * 10^2

So this means A = 2.63 and B = 2

6 0
3 years ago
Find the average rate of change of the function over the given intervals. f(x) = 12x3 + 12; [5,7]
AlladinOne [14]
<span>average rate of change = (f(7) - f(5)) / (x7 - x5)
</span>f(7) = <span>12 (7^3) + 12 = 4128
f(5) = </span>12 (5^3) + 12 = 1512

so
average rate of change = (4128 - 1512)/(7-5)
average rate of change = 2616 / 2
average rate of change = 1308

hope it helps
7 0
3 years ago
Which statement describes the graph of f(x)=-4x^4+3x^3+10x^2?
DedPeter [7]

Answer:

B is the answer, The graph touches the x axis at x = 0 and crosses the x axis at x = 5 and x = –2.

Step-by-step explanation:


8 0
3 years ago
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