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Thepotemich [5.8K]
3 years ago
11

What is the vertex of f(x)=x+8-3

Mathematics
2 answers:
kupik [55]3 years ago
4 0

Answer:

-2.5 I am not sure if it's the right answer

SpyIntel [72]3 years ago
3 0

Answer:

The absolute value vertex is  

( − 8 , − 3 ) .

Step-by-step explanation:

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What is an equation of the line that passes through the point ( − 5 , − 3 ) (−5,−3) and is parallel to the line 3 x + 5 y = 15 3
aivan3 [116]

Answer:

would it be 5y

Step-by-step explanation:

6 0
2 years ago
A toy store has six shells with five gang boxes on each shelf how many game boxes are thA toy store has six shells with five gan
docker41 [41]

Answer:

30 game boxes

Step-by-step explanation:

We have to multiply the number of each shelf by the number of game boxes,

The store has 6 shelves and 5 game boxes on each shelf.

Therefore, the number of game boxes that there are is:

6 * 5 = 30 game boxes

7 0
3 years ago
Is the square root of 48 rational or irrational?
Dmitrij [34]
Square root of 48= Square root of 16 times 3= 4 Square root of three. The square root of 48 is an irrational number!
7 0
3 years ago
Read 2 more answers
A building has a ramp to its front doors to accommodate the handicapped. If the distance from the building to the end of the ram
bija089 [108]
<span>The ramp forms a right triangle with legs 5 and 22. 
 Find the length of the hypotenuse using the Pythagorean theorem,
 c=raiz((5)^2+(22)^2)=</span><span> <span>22.56 ft
 Answer
</span></span> the ramp is 22.56 ft long. 
7 0
3 years ago
Which is the best approximation to a solution of the equation e^x = 2x + 3?
TiliK225 [7]
The correct question is 
Which is the best approximation to a solution of the equation
e^(2x) = 2e^{x) + 3?

we have that

e^(2x) = 2e^{x) + 3-----------> e^(2x)- 2e^{x) - 3=0
the term 
e^(2x)- 2e^{x)----------> (e^x)²-2e^(x)*(1)+1²-1²------> (e^x-1)²-1

then
e^(2x)- 2e^{x) - 3=0--------> (e^x-1)²-1-3=0------> (e^x-1)²=4
(e^x-1)=2--------> e^x=3
x*ln(e)=ln(3)---------> x=ln(3)
ln(3)=1.10
hence
x=1.10

the answer is x=1.10



3 0
3 years ago
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