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MissTica
3 years ago
13

Please I need help ..............................

Mathematics
1 answer:
erik [133]3 years ago
7 0

Answer:

You can type that info into M-a-t-h-w-a-y and they can give you a correct answer.

Step-by-step explanation:

Look it up without the dashes. this website wouldn't let me type it without for some reason. Hope this helps!

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1. A store priced a textbook at $126.75 using a markup of 34% of cost.
Alchen [17]
$43.09 of book
Shhebeenebbemansjdx
6 0
2 years ago
Someone plz help me plz
zheka24 [161]

Answer:

3(24+56)

Step-by-step explanation:

3 times sum of 24 and 56 so your adding 24 and 56 and multiplying it by 3

3(24+56)

8 0
3 years ago
Read 2 more answers
Find the value of sec(theta°)cos(theta°) for the following values of theta. <br> b. theta = 225
Nesterboy [21]

Answer:

sec(theta°)cos(theta°) = 1  

Step-by-step explanation:

given data

(theta°) = 225

to find out

sec(theta°)cos(theta°)

solution

as we know that given equation

(theta°) = 225

cos(theta°) will be

cos(225°) = -0.7071      .................................1

so we know

sec(theta°) = \frac{1}{cos(theta)}      ..............2

so put here value of cos(theta°)

sec(theta°) = \frac{1}{-0.7071}

sec(theta°) = - 1.4142

so

sec(theta°)cos(theta°) = -0.7071  × ( - 1.4142 )

sec(theta°)cos(theta°) = 1

so answer is sec(theta°)cos(theta°) = 1  

8 0
3 years ago
Simplify: m^2n^3 aanm^2a^2n^4
dolphi86 [110]

9514 1404 393

Answer:

  a^4·m^4·n^8

Step-by-step explanation:

The applicable rule of exponents is ...

  (a^b)(a^c) = a^(b+c)

__

m^2n^3 aanm^2a^2n^4

= a^(1+1+2)·m^(2+2)·n^(3+1+4)

= a^4·m^4·n^8

3 0
3 years ago
CAN SOMEONE PLZZ HELP ME THE LAST QUESTION PLZZZZZ?????
MissTica

first is proportional. y is 1 and 1/4 times x

second is not proportional.

reason? divided any y value by its corresponding x value and you'll keep getting 1.25, or 1 and 1/4. and the second one is not proportional because dividing the y values by their corresponding x value will give different answers.

maybe im wrong maybe im right you choose to trust me or not.

5 0
3 years ago
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