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inessss [21]
3 years ago
13

If log(a) = 1.2 and log(b)= 5.6, what is log(a/b)?

Mathematics
1 answer:
77julia77 [94]3 years ago
8 0

Answer:

d. -4.4​

Step-by-step explanation:

We know that log (a/b) = log a - log b

Since log a = 1.2 and log b = 5.6 , we can substitute these values into the equation.

                        log (a/b)  =   1.2 - 5.6

                                       = -4.4

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Rina8888 [55]

Given:

Pattern x: Starting number 5. Rule: Multiply by 3.

Pattern y: Starting number 20. Rule: Multiply by \dfrac{1}{2}.

To find:

The values for the given table.

Solution:

First value of x is 5.

The rule for pattern x is "number is multiply by 3".

Second value of x is:

5\times 3=15

Third value of x is:

15\times 3=45

The first value of y is 20.

The rule for pattern y is "number is multiply by \dfrac{1}{2}".

Second value of y is:

20\times \dfrac{1}{2}=10

Third value of y is:

10\times \dfrac{1}{2}=5

Therefore, the values in the table are 5, 15, 45 and the y-values are 20, 10, 5 respectively.

5 0
3 years ago
Given: ∆ABC ≅ ∆DEF<br> Find: AB and m∠F
Sindrei [870]

AB = 14

m∠F = 39°

Well the explanation here is simple.

Since the given states that the two triangles, ΔABC and ΔDEF, are congruent, they must have congruent/corresponding parts.

So AB is the same as DE which measures 14, so therefore AB must measure 14.

The same explanation goes to m∠F, ∠C is congruent to ∠F so it must measure 39° too.

3 0
3 years ago
5 - 3x &gt; -19 <br> Can someone please help me
oksano4ka [1.4K]

Answer:

- 3x > -19 -5

- 3x > -24

3x < 24

x < 8

8 0
3 years ago
One angle of a triangle measures 40°. The other two angles are in a ratio of 5:9. What are the measures of those two angles?
Travka [436]

Answer:

50 and 90

Step-by-step explanation:

(180-40)/(5 + 9) = 10

=> 50 and 90

5 0
3 years ago
There are 36 Red Marbles and 44 Blue marbles put into a bag. What is the probability of drawing a red marble out of the bag?
tigry1 [53]
The probability of drawing a red marble is 45%
I hope this was right and helpful and sorry for responding so late
7 0
3 years ago
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