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alexandr402 [8]
3 years ago
12

100x100x23x4x5 can somebody give me the it

Mathematics
1 answer:
devlian [24]3 years ago
4 0

Answer:

(100)(100)(23)(4)(5)

=4600000

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Solve for x.<br> -5/6x-1/9x=-102<br> x = ______?
kifflom [539]

Answer:x=1/108 is your answer

-5/6x-1/9x=-102

1/x(-5/6-1/9)=-102

1/x=-102/(-17/18)=-102×-18/17=108

x=1/108

3 0
3 years ago
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An angle with a measure of 180 degrees is formed by two __________ __________.
mario62 [17]
By two opposite rays. 

The explanation is t<span>he straight </span>angle<span> is the only type of </span>angle formed by two opposite rays<span>. All other </span>angles formed by two rays<span> are less than </span>180<span> degrees. Acute </span>angles<span> are between zero and 90 degrees. A right </span>angle<span>, with one </span>ray<span> perpendicular to another, is 90 degrees.

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3 years ago
Find the volume (show work)
Karo-lina-s [1.5K]

V = 1/3 × base × height

V = 1/3 × ( 8 × 8 ) × 7

V = 1/3 × 64 × 7

V = 448/3

V = 149.33333

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3 0
3 years ago
Find the number of positive integers less than 100,000 whose digits are among 1, 2, 3, and 4.
Paraphin [41]

As we can see that there are 6 digits in 100,000 and its is the smallest number we can have in 6 digit. So all numbers less than 100,000 will be 1-digit, 2-digits, 3-digits, 4-digits and 5-digits numbers made from 1,2,3,4 with repetitions allowed.

Case 1: All 1 -digit numbers

We will have numbers 1,2,3,4. So total 4 integers for this one

Case2: All 2-digit numbers

We can fill 1 digit place in 4 ways ( can choose any number out of 1,2,3,4). Then again we can fill 2nd digit place in 4 ways ( can choose any number out of 1,2,3,4). So all together we will have 4 × 4 = 16 integers for this one

Case3: All 3-digits numbers

We can fill 1 digit place in 4 ways ( can choose any number out of 1,2,3,4). Then again we can fill 2nd digit place in 4 ways ( can choose any number out of 1,2,3,4). similarly we can fill 3rd digit place in 4 ways (again any number out of 1,2,3,4). So all together we will have 4 × 4 × 4 = 64 integers for this one.

Case4: All 4-digit numbers

Again we can fill 1st digit place in 4 ways, then 2nd digit place in 4 ways, 3rd digit place in 4 ways, 4th digit place in 4 ways. So all together there will be 4 × 4 × 4 × 4 = 256 integers for this one.

Case5: All 5-digit numbers

Again we can fill 1st digit place in 4 ways, then 2nd digit place in 4 ways, 3rd digit place in 4 ways, 4th digit place in 4 ways, 5th digit place also in 4 ways. So all together there will be 4 × 4 × 4 × 4 × 4 = 1024 integers for this one

Adding results of all 5 cases we get,

Total integers = 4 + 16 +64 + 256 + 1024 = 1364 integers.

So thats the final answer

4 0
3 years ago
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gogolik [260]

Step-by-step explanation:

8 log_{5}( {x}^{5} )  = 8 \times 5 log_{5}(x)  \\  = 40 log_{5}(x)

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