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Luden [163]
3 years ago
13

Help with geometry !!!!!!!!!!!

Mathematics
2 answers:
My name is Ann [436]3 years ago
7 0

Answer

x = 18

You're welcome

irinina [24]3 years ago
7 0

Answer:

The answer would be x=18

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Math models helpppp plss if you know about math models answer this pls
svet-max [94.6K]

Answer:

A mathematical model is a description of a system using mathematical concepts and language. The process of developing a mathematical model is termed mathematical modeling. Mathematical models are used in the natural sciences and engineering disciplines, as well as in non-physical systems such as the social sciences.

4 0
3 years ago
The first step to finding the quotient is to ?
guajiro [1.7K]

Answer:

divide one number by another. dividend ÷ divisor = quotient.

Step-by-step explanation:

Example: in 12 ÷ 3 = 4, 4 is the quotient.

5 0
3 years ago
If an object is dropped from a height of 200 feet, the function h(t)=-16t^2=200 gives the height of the object after t seconds.
Fed [463]
Hello,
Answer C
h(t)=0=-16t²+200==>t=3.53533...(s)
5 0
3 years ago
Initially, there were only 86 weeds in the garden. The weeds grew at a rate of 29% each week. The following function represents
Orlov [11]

Step-by-step answer:

The base of the exponential function is 1.29 for 7 days, as in

f(x) = 86*(1.29)^x

The new rate for days can be calculated by dividing x by 7 (where x remains the number of weeks), namely

f(x) = 86*1.29^(x/7)

Using the law of exponents, b^(x/a) = b^(x*(1/a)) = (b^(1/a))^x

we simplify by putting b=1.29, a=7 to get

f(x) = 86*(1.29^(1/7))^x

f(x) = 86*(1.037)^x      since 1.29^(1/7) evaluates to 1.037

Rounding 1.037 to 1.04 we get a (VERY) approximate function

f(x) = 86 * (1.04^x)

1.04 is very approximate because 1.04^7 is supposed to get back 1.29, but it is actually 1.316, while 1.037^7 gives 1.2896, much closer to 1.29.

7 0
3 years ago
Read 2 more answers
How can you the unit rate on a graph that goes through the origin ?
Nonamiya [84]
It must be about the origin in order to do that
5 0
3 years ago
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