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ser-zykov [4K]
3 years ago
14

Write the algebraic expression that matches each graph? PLS HELP I NEED HELP!!!

Mathematics
2 answers:
Vladimir [108]3 years ago
8 0

Looks like y = |x| because the slopes of the lines are both 1

You shift it to the right making it |x-2| and you shift it down making y = -2

combine to get <em>y = |x - 2| - 2</em>.

vlada-n [284]3 years ago
3 0

Answer: -/x/-3

Step-by-step explanation:

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Slope= -1 /5<br> Y-Intercept = -3 <br>Point-Slope Form:​
Fittoniya [83]

Answer:

3

Step-by-step explanation:

4 0
2 years ago
Answer plzzz I’ll give u more points
Olenka [21]
13. $9 because 75%=3/4 and 1/4 of 12 is 3, and so you multiply $3 by 3 to get $9
14. Multiply 325 by .2 to get $65.00
15. 15 minutes, because if it rises 2 feet every hour, then it rises 1 foot every hour, and so you divide that by 2 to get 15
Answers:
13. $9
14. $65.00
15. 15 mins
4 0
3 years ago
How do i do this? Its confusing
cestrela7 [59]

Answer:

$12,415.48

Step-by-step explanation:

A = P (1 + r/n)^(nt)

where A is the final amount,

P is the initial amount,

r is the annual interest rate as a decimal,

n is the number of compoundings per year,

and t is the number of years.

A = 8000 (1 + 0.152/2)^(2×3)

A = 8000 (1.076)^6

A = 12415.48

5 0
3 years ago
How to solve -d/6+12=-7
Delicious77 [7]
IF you are solving for d:

isolate the D, do the opposite of PEMDAS.

-d/6 + 12 = -7

(subtract 12 from both sides)

-d/6 + 12 (-12) = -7 (-12)

-d/6 = -19

(multiply 6 to both sides)

-d/6(6) = -19(6)

-d = -19(6)

-d = -144

-d/-1 = -144/-1

d = 144

hope this helps
6 0
3 years ago
A trough of water is 20 meters in length and its ends are in the shape of an isosceles triangle whose width is 7 meters and heig
Vaselesa [24]

Answer:

a) Depth changing rate of change is 0.24m/min, When the water is 6 meters deep

b) The width of the top of the water is changing at a rate of 0.17m/min, When the water is 6 meters deep

Step-by-step explanation:

As we can see in the attachment part II, there are similar triangles, so we have the following relation between them \frac{3.5}{10} =\frac{a}{h}, then a=0.35h.

a) As we have that volume is V=\frac{1}{2} 2ahL=ahL, then V=(0.35h^{2})L, so we can derivate it \frac{dV}{dt}=2(0.35h)L\frac{dh}{dt} due to the chain rule, then we clean this expression for \frac{dh}{dt}=\frac{1}{0.7hL}\frac{dV}{dt} and compute with the knowns \frac{dh}{dt}=\frac{1}{0.7(6m)(20m)}2m^{3}/min=0.24m/min, is the depth changing rate of change when the water is 6 meters deep.

b) As the width of the top is 2a=0.7h, we can derivate it and obtain \frac{da}{dt}=0.7\frac{dh}{dt}  =0.7*0.24m/min=0.17m/min The width of the top of the water is changing, When the water is 6 meters deep at this rate

8 0
3 years ago
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