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3 years ago

Put in slope intercept form

Mathematics

1 answer:

LiRa [457]3 years ago

3 0

These require you solve for y. Do that by subtracting the x-term, then dividing by the y-coefficient.

5. 3y = 7x -9
y = (7/3)x -3

6. 6y = -2x -3
y = (-2/6)x -3/6
y = (-1/3)x -1/2

7. 8y = 2x + 12
y = (1/4)x +3/2

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Choose all of the addition Adams in which you in exams to align place values in the Attic

solong [7]

+, //, attic. I think this is what your asking be more detailed.

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2 years ago

What is the value of a? A) 28 B) 45 C) 62 D) 73

Veronika [31]

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2 years ago

Read 2 more answers

A small metal bar, whose initial temperature was 10° C, is dropped into a large container of boiling water. How long will it tak

IgorC [24]

Answer

According to newton law of cooling

$\dfrac{d T}{dt} = k (T - T_a)$

and

$T = Ce^{kt} + T_a$

now At Ta = 100

T₀ = 10 °C

T₁ = 12° C

$T(0) = Ce^{0} + T_a$

10 = C + 100

C = -90

now,

$12 = -90e^k + 100$

$e^k = \dfrac{88}{90}$

$k = ln{\dfrac{88}{90}}$

at time equal to t

T(t) = 70

$T(t) = -90 (\dfrac{88}{90})^t + 100$

$70 = -90(\dfrac{88}{90})^t + 100$

$(\dfrac{88}{90})^t = \dfrac{1}{3}$

$t\times ln(\dfrac{88}{90}) = ln(\dfrac{1}{3})$

t = 48.88 s

hence, after 48.88 s the temperature of the body will be 70°C

b) time taken to reach 98°C

$T(t) = -90 (\dfrac{88}{90})^t + 100$

$98 = -90(\dfrac{88}{90})^t + 100$

$(\dfrac{88}{90})^t = \dfrac{1}{45}$

$t\times ln(\dfrac{88}{90}) = ln(\dfrac{1}{45})$

t = 390 s

hence, after 390 s the temperature of the body will be 98°C

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2 years ago

Convert 6% to an equivalent decimal

Burka [1]

Answer:

0.06

Step-by-step explanation:

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2 years ago

Differentiate. F(x) = (x3 - 3)2/3 2x f'(x) - 3 8 O f'(x) = 3 8 2x2 f'(x) 3 VX3 -8 f(x) 3 V3-8

oee [108]

Answer:

$f'(x) = \frac{2{x}^{2}}{ \sqrt[3]{( {x}^{3} - 8) } }$

Step-by-step explanation:

$f(x) = ( {x}^{3} - 8)^{ \frac{2}{3} } \\ \\ f'(x) = \frac{2}{3} ( {x}^{3} - 8)^{ \frac{2}{3} - 1 } (3 {x}^{2} - 0) \\ \\ f'(x) = \frac{2}{3} ( {x}^{3} - 8)^{ \frac{2 - 3}{3} } \times 3 {x}^{2} \\ \\ f'(x) = 2{x}^{2}( {x}^{3} - 8)^{ \frac{ - 1}{3} } \\ \\ f'(x) = \frac{2{x}^{2}}{( {x}^{3} - 8)^{ \frac{ 1}{3} } } \\ \\ \huge \red{ \boxed{ f'(x) = \frac{2{x}^{2}}{ \sqrt[3]{( {x}^{3} - 8) } } }}$

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2 years ago