Greatest common multiple of 3 and 4

Please help me with the following questions. Thanks in advance!!

Orlov [11]

Answer:

Step-by-step explanation:

Eight

590 = 1800 e^(-0.4t) Divide by 1800

0.32778 = e^(-0.4t) Put the value for e in the denominator.

0.32778 = 1 / e^(0.4t) Notice the sign's gone. Multiply both sides by e^(0.4t)

0.32778 *e^(0.4t) = 1 Divide by 0.32778

e^(0.4t) = 3.0508 Take the natural log of both sides.

0.4t = 1.11542 Divide both sides by 0.4

t = 2.78855

You should put this in the original equation to see if it checks. It does.

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Nine

3^(3 - 3x) + 6 = 42 Subtract 6 from both sides

3^(3 - 3x) = 36 Take the log of both sides.

(3 - 3x) ln(3) = ln 36

(3 - 3x) *1.098612 = 3.58352 Divide by 1.098612

3 - 3x = 3.26186 Subtract 3 from both sides

- 3x = 0.26186 Divide by - 3

x = - 0.087287

4 0

3 years ago

Write the equation of the line that passes through each pair of points in slope-intercept form. This is worth 3 points. You have

LekaFEV [45]

(0,5)(3,-1)
slope = (-1 - 5) / (3 - 0) = -6/3 = -2

y = mx + b
slope(m) = -2
(use either of ur points...(0,5)...x = 0 and y = 5
now we sub and find b, the y int
5 = -2(0) + b
5 = b

so ur line is : y = -2x + 5

6 0

2 years ago

Read 2 more answers

Please help me idk this

Soloha48 [4]

Answer:

round it up and 8 is closer to ten so it would be 4 pounds

4 0

3 years ago

Read 2 more answers

If the original square had a side length of

irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$