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katovenus [111]
4 years ago
10

How can I condense this expression into a single logarithm?

Mathematics
1 answer:
Sholpan [36]4 years ago
4 0
Because they have the same base you can When you subtract logs of the same base you divide them Aka log5-log4= log(5/4) Similarly when you add you multiply Aka log5+log4= log (5*4) Make sure they have the same base like in your problem
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Rectangle JKLM:J(-3, 8), K(-3,-1), L(4, -1),
Maslowich

Answer:

The perimeter of the Rectangle JKLM is 32 unit.

Step-by-step explanation:

Consider the provided coordinates.

Rectangle JKLM: J(-3, 8), K(-3,-1), L(4, -1),  M(4,8)

First plot the points as shown in figure:

The required figure is shown below:

Now find the distance between JK by using the distance formula.

d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

d=\sqrt{\left(-3-\left(-3\right)\right)^2+\left(-1-8\right)^2}=9

Therefore, the length of JK is 9 unit.

Now find the length of the side JM

d=\sqrt{\left(4-\left(-3\right)\right)^2+\left(8-8\right)^2}=7

Therefore, the length of JM is 7 unit.

The perimeter of rectangle is: P=2(l+b)

Therefore,

P=2(7+9)

P=2(16)

P=32

Hence, the perimeter of the Rectangle JKLM is 32 unit.

4 0
4 years ago
When constructing an equilateral triangle by hand, which step comes after constructing a circle?
KonstantinChe [14]

Answer:

set compass to the radius of the circle

Step-by-step explanation: when constructing an equilateral triangle by hand . circle constructing by set compass to the radius of the circle and rotate it and come to the initial point

4 0
3 years ago
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Given: FH ⊥ GH; KJ ⊥ GJ Prove: ΔFHG ~ ΔKJG Identify the steps that complete the proof. ♣ = ♦ = ♠ =
Feliz [49]

1 is all right angles are congruent 2 angles fgh is congruent to angle kgj 3 aa similarity theorem just took the test this are correct

6 0
3 years ago
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A circle is drawn within a square as shown. What is the best approximation for the area of the shaded region? Use 3.14 to approx
fgiga [73]
In the shown picture, To find the shaded area we need to find the total area of the square, subtracted by the area of the circle.

the area of the shaded region = the area of the square - the area of the circle

the area of the square = L² = 7² = 49 cm²

the area of the circle = πr² = 3.14 x 3.5² = 38.5 cm²

the area of the shaded region = 49 - 38.5 = 10.5 cm²

3 0
3 years ago
I need help (last question)
svet-max [94.6K]
What’s the last question? please attach a photo
7 0
3 years ago
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