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givi [52]
3 years ago
9

Solve the equation. 4k^2 = 5k

Mathematics
2 answers:
stich3 [128]3 years ago
8 0
The answer is d. 4(1)^2=5(4/5). 4=4.
Mekhanik [1.2K]3 years ago
3 0
The answer is B. Your going to subtract 4k^2 from both sides, which it would be 4k^2-5k=0. Then you going to factor k(4K-5)=0, which k=0 and 4K-5=0 and then k =0 and 4/5
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Can someone help me w questions 1 and 2 i’ll give u brainliest
Leni [432]

Answer:

1. x^2+5

2.x^2-6x+9

Step-by-step explanation:

these are just translations up and to the right.

since the f(x)=x^2

the transformations would be

1. g(x)=x^2+5

2.g(x)=(x-3)^2 = x^2-6x+9

4 0
2 years ago
Can someone help with this question
IrinaK [193]

210 - ( 30 * x )

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6 0
3 years ago
Is 0.529 greater than 0.54
attashe74 [19]
No. 0.54 is equal to 0.540

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6 0
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Read 2 more answers
10. The sum of two numbers is 13 more than twice the first number. Their difference is 14 less than the second number. Find each
Lynna [10]

The sum of two numbers is 13 more than twice the first number:

x + y = 2x +13

x - y = y - 14

solve the system of equations:

x - (x + 13) = (x +13) - 14

x = -12

y = 1

   

3 0
3 years ago
A pizza pan is removed at 9:00 PM from an oven whose temperature is fixed at 450°F into a room that is a constant 70°F. After 5​
gladu [14]

Answer:

A) It will get to a temperature of 125°F at 9:19 PM

B) It will get to a temperature of 150°F at 9:16 PM

C) as time passes temperature approaches the initial temperature of 450°F

Step-by-step explanation:

We are given;

Initial temperature; T_i = 450°F

Room temperature; T_r = 70°F

From Newton's law of cooling, temperature after time (t) is given as;

T(t) = T_r + (T_i - T_r)e^(-kt)

Where k is cooling rate and t is time after the initial temperature.

Now, we are told that After 5​ minutes, the temperature is 300°F.

Thus;

300 = 70 + (450 - 70)e^(-5k)

300 - 70 = 380e^(-5k)

230/380 = e^(-5k)

e^(-5k) = 0.6053

-5k = In 0.6053

-5k = -0.502

k = 0.502/5

k = 0.1004 /min

A) Thus, at temperature of 125°F, we can find the time from;

125 = 70 + (450 - 70)e^(-0.1004t)

125 - 70 = 380e^(-0.1004t)

55/380 = e^(-0.1004t)

In (55/380) = -0.1004t

-0.1004t = -1.9328

t = 1.9328/0.1018

t ≈ 19 minutes

Thus, it will get to a temperature of 125°F at 9:19 PM

B) Thus, at temperature of 150°F, we can find the time from;

150 = 70 + (450 - 70)e^(-0.1004t)

150 - 70 = 380e^(-0.1004t)

80/380 = e^(-0.1004t)

In (80/380) = -0.1004t

-0.1004t = -1.5581

t = 1.5581/0.1004

t ≈ 16 minutes.

Thus, it will get to a temperature of 150°F at 9:16 PM

C) As time passes which means as it approaches to infinity, it means that e^(-kt) gets to 1.

Thus,we have;

T(t) = T_r + (T_i - T_r)

T_r will cancel out to give;

T(t) = T_i

Thus, as time passes temperature approaches the initial temperature of 450°F

6 0
3 years ago
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