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3 years ago

What is the solution for x in the equation? -2x + 14 + 10x = 34 A. X=6 B. X=2/5 C. X=1/8 D. X=5/2

Mathematics

1 answer:

julsineya [31]3 years ago

5 0

Answer:

Step-by-step explanation:

x =(10-√-800)/2=5-10i√ 2 = 5.0000-14.1421i

x =(10+√-800)/2=5+10i√ 2 = 5.0000+14.1421i

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Triangle HIJ has vertices H(1, 7) , I(−4, −2) , and J(−3, 5) . Write the coordinate notation for a translation of 3 units right

e-lub [12.9K]

Step-by-step explanation:

(x, y) -> (x+3, y+4)

that is what 3 units to the right (3 units into the outsource x direction) and 4 units up (4 units into the posits y directing) mean.

so, all points go through this translation

(1, 7) -> (4, 11)

(-4, -2) -> (-1, 2)

(-3, 5) -> (0, 9)

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2 years ago

A package of 5 crackers contains 205 Calories. How many Calories are in one cracker using a double number line

natulia [17]

Answer:

Step-by-step explanation:

since there are 205 calories in 5 crackers we can represent this as

205/5

we need to find how many calories are in one cracker, so lets take "x" as the # of calories in 1 cracker

$\frac{x}{1} =\frac{205}{5}$

cross multiply

5x=205

divide by 5 on both sides

x=41

41 calories in 1 cracker

Number line:

7 0

3 years ago

This is a dumb question but what is 68 time 100

Whitepunk [10]

68 x 100 = 6,800

Hope this helps

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3 years ago

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Explain how solving 8+2 can help you solve 80+20.

katrin [286]

Becuase 80 is the same as 8 but has an extra 0 at the end (which makes it 80) and same for 20 to 2. 8 + 2 = 10 (now add a zero at the end). The answer should be 100.

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3 years ago

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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 =

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

4 0

3 years ago