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2 years ago

Identify all terms in the expression 5x³ + 4x² + 3x + 1

Mathematics

1 answer:

lorasvet [3.4K]2 years ago

7 0

The answer to your question is 145x

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-4x + 7(x + (-2)) > -18

professor190 [17]

-4x+7x-14
3x-14>-18
3x>-4
X>-4/3

3 0

3 years ago

Michelle drew a scale drawing of a swimming pool she used the scale 14 millimeters 5 meters in the drawing the pool is 42 millim

Lera25 [3.4K]

EXPLANATION

Let's see the facts:

The scale is---> 14 milimeters ----> 5 meters

Width = 42 milimeters

Applying the unitary method, the actual width of the pool is:

$Actual_{}width=42\text{ mm}\cdot\frac{5m}{14\text{ mm}}=15m$

The width of the pool is 15m

6 0

6 months ago

Zohar is using scissors to cut a rectangle with a length of 5x – 2 and a width of 3x + 1 out of a larger piece of paper. The per

Tresset [83]

62 is the answer, 2(5(4)-2)+2(3(4)+1)=62

8 0

2 years ago

Read 2 more answers

Describe the end behavior of the given function <br> f(x)=9x^3-x^2-8x+6<br>

Mekhanik [1.2K]

As x approaches -inf f(x) -> -inf
and as x approaches inf, f(x) approaches +inf

Mark brainliest please

6 0

2 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

2 years ago