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rosijanka [135]
2 years ago
12

In the last quarter of a high school football game, your team i s behind by 21 points. A field goal is worth 3 points and a touc

hdown (after the extra point is worth 7 points. Let x represent the number of field goals scored and y represent the number of touchdowns scored. Write an inequality that models the different number of field goals and touchdowns your team could score to win the game. (Assume the other team doesn't score any more points.)​
Mathematics
2 answers:
Cerrena [4.2K]2 years ago
4 0

Answer:

3x + 7y > 21

Step-by-step explanation:

3x + 7y > 21

where:

x = number of field goals needed to win

y = number of touchdowns (plus extra point) needed to win

Since your team is 21 points behind and you want to win the game, not tie it, the number of field goals and touchdowns must result in a number higher than 21, not equal to or higher than.

MAXImum [283]2 years ago
3 0

Answer:

3x + 7y > 21

3x + 7y > 21

where:

x = number of field goals needed to win

y = number of touchdowns (plus extra point) needed to win

Since your team is 21 points behind and you want to win the game, not tie it, the number of field goals and touchdowns must result in a number higher than 21, not equal to or higher than.

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Use compatible numbers to estimate the quotient.674.8 ÷ 23
-BARSIC- [3]

Answer:

~30

Step-by-step explanation:

By using division, the answer is 29.34. If the question is asking to estimate, I would estimate the quotient as 30.

5 0
3 years ago
A. When comparing the "# of Bacteria" each hour, what is the number being multiplied by each time? *
LenaWriter [7]

Answer:

Part a) The number is 8

Part b) The initial number of bacteria is 4

Part c) y=4(8)^x

Part d) y=100(8)^x

Step-by-step explanation:

we know that

The equation of a exponential function is equal to

y=a(b)^x

where

y is the number of bacteria

x is the time in hours

b is the base of the exponential function

a is the initial number of bacteria

Part a) When comparing the "# of Bacteria" each hour, what is the number being multiplied by each time?

we know that

For x=1 h ----> y=32 bacteria

For x=2 h ----> y=256 bacteria

For x=3 h ----> y=2,048 bacteria

For x=4 h ----> y=16,384 bacteria

For x=5 h ----> y=131,072 bacteria

For x=6 h ----> y=1,048,576 bacteria

so

256/32=8

2,048/256=8

16,384\2,048=8

131,072/16,384=8

1,048,576\131,072=8

so

the base of the exponential function  b is 8

Part b) What is the initial number of bacteria? (at time zero)

we know that

The number of bacteria at time x=1 hour , divided by the number of bacteria at time x=0 must be equal to 8 (see part a)

so

\frac{32}{a}=8

solve for a

a=32/8=4\ bacteria

Part c) Write a rule for this table

we have

y=a(b)^x

we have

a=4\\b=8

substitute

y=4(8)^x

Part d) Suppose you started with 100 bacteria, but they still grew by the same growth factor. Write the function rule for this situation

In this case

a=100

b=8 ---> the growth factor is the same

so

y=100(8)^x

6 0
3 years ago
Read 2 more answers
A bottle contains 568 millilitres of milk, jack pours out half a litre. How much milk is left
aleksandrvk [35]
284 which means 568 divide 2 gives u 284
3 0
3 years ago
Tonisha has a lemonade stand. She has $25 in expenses and wants to make at least $55 per day.
Novosadov [1.4K]

Hello!

Your answer is:

x-31 ≥ 66⇒x ≥ 97!

Hopefully this helps! Please correct me if im wrong! :C

5 0
3 years ago
If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=
Alexxx [7]
\bf f(x)=y=2x+sin(x)
\\\\\\
inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
\\\\\\
\textit{now, the "y" in the inverse, is really just g(x)}
\\\\\\
\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
-----------------------------\\\\

\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
\\\\\\
1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}

now, if we just knew what g(2)  is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that  means if we use that value in f(x),   f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus    2 = 2x+sin(x)

\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2
\\\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
5 0
2 years ago
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