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dsp73
3 years ago
13

How bigger is a 1 ¼ inch button than a ⅝ inch button

Mathematics
1 answer:
Harrizon [31]3 years ago
8 0
I believe the answer is 5/8 
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A golf course charges $16 for a package including the full 18-hole course. The course also sells buckets of
MA_775_DIABLO [31]

16x + 21y = 555

Step-by-step explanation:

Let x be the no. of 18-hole course

And y be the no. of golf balls

16x + 21y = 555

6 0
3 years ago
What is the measure of angle DBC
MAXImum [283]
The angle measure of DBC is 45
3 0
3 years ago
Estimate then find the difference or sum <br><br> 86+17
kumpel [21]
Estimation for addition: 86 rounds up to 90 and 17 rounds up to 20 so:
90 + 20 = 110
Estimation for subtraction is the same thing but with subtraction
90 - 20 = 70

Now for the exact answers:
86 + 17 = 103
86 - 17 = 69

Your answers are 110, 70, 103, and 69.

8 0
3 years ago
A closed container has 5.04 ⋅ 1023 atoms of a gas. Each atom of the gas weighs 1.67 ⋅ 10−24 grams.
MatroZZZ [7]

Answer:

The correct option is the last one (0,84 grams, because (5.04 × 1.67) × (10^23 x 10^-24) = 8.4168 x 10^-1))

Step-by-step explanation:

The first step is to think that you have to multiply the number of atoms for the weight of each atom.

When you work with scientific notation, you should solve the operation for numbers and for the ten base. In this case you solve 5.04 x 1.67 and then the product of the ten bases. For this product you have to sum the exponents of the ten base 23 + (-24) = -1.

Summarizing.... 5.04 x 10^23 x 1.67 x 10^-24 = (5.04 x 1.67) x (10^23x10^-24)  ⇒ 5.04 x 10^23 x 1.67 x 10^-24 = 8.4168 x 10^-1

⇒ 5.04 x 10^23 x 1.67 x 10^-24 = 0,84168.

4 0
3 years ago
Given ACM, angle C=90º. AP=9, PM=12. Find AC, CM, AM.
Gnesinka [82]

Answer:

AM = 25, AC = 15, CM = 20

Step-by-step explanation:

The given parameters are;

In ΔACM, ∠C = 90°, \overline{CP} ⊥ \overline{AM}, AP = 9, and PM = 16

\overline{AC}² + \overline{CM}² = \overline{AM}²

\overline{AM} = \overline{AP} + PM = 9 + 16 = 25

\overline{AM} = 25

\overline{AC}² = \overline{AP}² + \overline{CP}² = 9² +  \overline{CP}²

∴ \overline{AC}² = 9² +  \overline{CP}²

Similarly we get;

\overline{CM}² = 16² + \overline{CP}²

Therefore, we get;

\overline{AC}² + \overline{CM}² = 9² +  \overline{CP}² + 16² + \overline{CP}² = \overline{AM}² = 25²

2·\overline{CP}² = 25² - (9² + 16²) = 288

\overline{CP}² = 288/2 = 144

\overline{CP} = √144 = 12

From \overline{AC}² = 9² +  \overline{CP}², we get

\overline{AC} = √(9² +  12²) = 15

\overline{AC} = 15

From, \overline{CM}² = 16² + \overline{CP}², we get;

\overline{CM} = √(16² + 12²) = 20

\overline{CM} = 20.

3 0
3 years ago
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