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Nostrana [21]
3 years ago
8

Please help thx ______________

Mathematics
2 answers:
lubasha [3.4K]3 years ago
7 0

Answer:

She drew a rectangle

Step-by-step explanation:

Rectangles have four right angles, and two pairs of (opposite) congruent sides.

zheka24 [161]3 years ago
7 0
Rectangle because the sides would be straight due to the 90 degree angle and sides would be parallel
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Xavier rides his bike 5 1/4 miles or 82% of his goal for the day. What is his goal for the day?
Leno4ka [110]
Use ratio and proportion

5.25/x= 82/100
cross multiply
82 x= 525
     x= 6.4 miles

Hope this helps
8 0
3 years ago
A coin contains 1.25 grams of nickel and 3.75 grams of copper, for a total weight of 5 grams. What percentage of the metal in th
otez555 [7]
So,

To find the percentage that the copper comprises, just divide the amount of copper by the total amount of metal in the coin.  In this case, we will change the denominator so that it is 100, which can easily be converted into a percentage, which is really just a fraction out of 100.

\frac{3.75}{5} = \frac{7.50}{10} = \frac{75}{100}

75%

75% of the metal in the coin is copper.
7 0
3 years ago
If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(
timama [110]

Use the change-of-basis identity,

\log_x(y) = \dfrac{\ln(y)}{\ln(x)}

to write

xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}

Use the product-to-sum identity,

\log_x(yz) = \log_x(y) + \log_x(z)

to write

xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}

Redistribute the factors on the left side as

xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}

and simplify to

xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)

Now expand the right side:

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}

Simplify and rewrite using the logarithm properties mentioned earlier.

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1

xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}

xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}

xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)

\implies \boxed{xyz = x + y + z + 2}

(C)

6 0
2 years ago
The identification code on a bank card consists of 1 digit followed by 2 letters. The code must meet the following conditions: T
jasenka [17]

Answer:

For the code we have 3 selections.

The first selection is a digit that must be odd, so the options are {1, 3, 5, 7 ,9}

So we have 5 options.

The second selection is a letter from the set of all the letters (27) minus the set of the vowels (5)

So here we have 27 - 5 = 22 options

The third selection is also a letter from the previous set, but because each letter can be used only one time, and in the previous selection we already selected one of the letters, in this selection we have a letter less than in the previous selection.

Here we have 22 - 1 = 21 options.

The total number of combinations (of possible codes) is equal to the product of the number of options for each selection:

C = 5*22*21 = 2310.

There are 2310 different possible codes

8 0
3 years ago
Try it, let's see can u do or not
Firlakuza [10]

Answer:

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
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