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Yuki888 [10]
3 years ago
14

What algebraic expression is 5+3(2x-7)=5+6x-21

Mathematics
1 answer:
Nesterboy [21]3 years ago
7 0
I am hundred percent sure yur answer is 10

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Given that Ax - 4x² + 3x² + 4 = (x² - 2x + 1)(2x² + B) + C(x-1) + D for all values of x, evaluate A, B, C and D. Hence, deduce t
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2 years ago
A rectangular poster is 5 dm long and 3 dm wide. Sara wants to paint the whole poster. Find the area of the poster in square cm.
Zina [86]

Answer:

\boxed{1500cm^{2} }

Step-by-step explanation:

Area = Length × Width

→ Identify the length and width

Length = 5 and Width = 3

→ Substitute in the values

Area = 5 dm × 3 dm

→ Convert into cm

Area = 50 cm × 30 cm

→ Simplify

1500 cm²

6 0
2 years ago
Are the triangles congruent if they are which theorem and why?
leonid [27]
Yes the sides are equal to each other
7 0
3 years ago
To find the quotient 3/4 divided by 1/8
vfiekz [6]

Answer:

B

Step-by-step explanation:

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
2 years ago
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