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Leviafan [203]
3 years ago
15

Hey please help me be geniune this is a timed test and I really need help will mark brianliest :( please no links I’m tired of s

eeing them

Mathematics
1 answer:
kumpel [21]3 years ago
8 0

Answer:

510 ft^2

Step-by-step explanation:

(12*9)+(12*9)+(7*9)+(7*9)+(7*12)+(7*12) = 108+108+63+63+84+84 = 510 ft^2

Hope this helped! :)

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How to write proofs in geometry
sp2606 [1]
Make a game plan. ...
Make up numbers for segments and angles. ...
Look for congruent triangles (and keep CPCTC in mind). ...
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Look for parallel lines. ...
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Use all the givens.
5 0
3 years ago
if you needed only 1c of milk, what is your best choice at the grocery store- a quart container, a pint container, or a 1/2 gal
Artist 52 [7]
A pint container because there are 2 cups in a pint, whereas there are 6 cups in 1/2 gallon and 4 cups in a quart.
3 0
3 years ago
Read 2 more answers
PLEASE HELP WITH ALGEBRA QUESTION NEED ASAP!!
Delvig [45]

Answer:

The function f(x) = ln(x - 4) is graphed the question options

Step-by-step explanation:

* Lets study the the information of the problem

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* That means the curve gets closer and closer to the vertical line x = 4

  but does not cross it

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* That means if we substitute x = e + 4 in the equation the value

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- It has an x-intercept of 5

* That means if we substitute y = 0 in the equation the value of x

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* Lets find the right answer

∵ f(x) = ln(x - 4)

* To find the equation of the asymptote let x - 4 = 0

∵ x - 4 = 0

∴ x = 4

∴ f(x) has a vertical asymptote at x = 4

* Lets check the point (e + 4 , 1) lies on the graph of the f(x)

∵ x = e + 4

∴ f(e+4) = ln(e + 4 - 4) = ln(e)

∵ ln(e) = 1

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* Change the logarithmic function to the exponential function

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∴ x - 4 = 1 ⇒ add 4 to the both sides

∴ x = 5

* The function f(x) = ln(x - 4) is graphed the question options

3 0
3 years ago
What is the domain of f(x) = 3x?
Stolb23 [73]
Answer is C.
All real numbers

hope it helps
5 0
3 years ago
Read 2 more answers
PLEASE HELP!!!!!!!!!!!
Art [367]

green pepper, onion, tomato

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6 0
3 years ago
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