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3 years ago

Algebra help please I need a equation please

Mathematics

1 answer:

Rudik [331]3 years ago

3 0

K = s + 4
k + s = 26
s + 4 = 26
Subtract 4 from both sides of the equation.
s = 22
Steve ran 22 miles.

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Standard algorithm for multiplication

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A standard algorithm is a step-by-step way to solve a problem.

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3 years ago

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(7x-49x^2)/(7x) can smeone solve this? Thank you! I hope you have a great day!

svp [43]

Answer:

x = 1/7

Step-by-step explanation:

$\frac{(7x-49x^2)}{7x}\\\\\frac{7x(1-7x)}{7x} \\\\1-7x=0\\\\1=7x\\\\$

x = 1/7

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3 years ago

BRAINLIEST For the fastest answer

MaRussiya [10]

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the answer is b wait I can also be c and d

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2 years ago

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if the first and third of three consecutive odd integers are added, the result is 63 less than five times the second integer. fi

ollegr [7]

The value of third consecutive odd is 23

• In Mathematics, algebra is a branch which only deals with symbols and the rule for the manipulation of these symbols. Algebra is divided into different sub branches such as advanced algebra, linear algebra, elementary algebra, abstract algebra etc.

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Let the first consecutive odd integer be x

Second consecutive integer = x+2

Third consecutive integer = x+4

We are given that

First + third = 5(Second) – 63

x + x + 4 = 5 (x+2) -63

2x + 4 = 5x +10 -63

-3x = -57

3x = 57

x = 19

First Consecutive odd = 19

Second Consecutive odd = 21

Third Consecutive odd = 23

Learn more about algebra here

brainly.com/question/723406

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2 years ago

Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be

Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C' (C' = kC)

taking exponents of both sides, we have

$L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}$

When t = 0, A(0) = 0 (since the forest floor is initially clear)

$A = \frac{L}{k} - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k} - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k} - \frac{C"}{k} e^{0}\\\frac{L}{k} = \frac{C"}{k} \\C" = L$

$A = \frac{L}{k} - \frac{L}{k} e^{kt}$

So, D = R - A =

$D = \frac{L}{k} - \frac{L}{k} - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}$

when t = 0(at initial time), the initial value of D =

$D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}$

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3 years ago