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lisabon 2012 [21]
3 years ago
13

The length of a photograph is 3cm less than twice the width. The area is 77 cm. find the dimensions of the photograph

Mathematics
1 answer:
Mila [183]3 years ago
4 0
I think it is 25.6 or something around there
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Step-by-step explanation:

it sound goos

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(picture)<br>complete it​
Free_Kalibri [48]

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Step-by-step explanation:

I don't have a good idea

8 0
3 years ago
Read 2 more answers
Why is (-3,6] a better answer than (-3,6)
Pani-rosa [81]

(-3, 6) includes all numbers between -3 and 6 but it does not include -3 and 6.

when you put square parentheses you include the number on the edge

so if an anwser includes all numbers between -3 and 6 and the number 6 it's correct to write (-3,6]

7 0
2 years ago
Help please!!! I dont understand these questions<br><br><br>currently attaching photos dont delete
Katyanochek1 [597]

Answer:

  1. b/a
  2. 16a²b²
  3. n¹⁰/(16m⁶)
  4. y⁸/x¹⁰
  5. m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

__

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}

__

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2

__

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}

__

4. This works the same way the previous problem does.

\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}

__

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}

3 0
3 years ago
Simplify 7y+2x - 10y+3x to the second power -10x
lara [203]

7y+2x - 10y+3x^2-10x=(3x^2)+(2x-10x)+(7y-10y)\\\\=3x^2-8x-3y

4 0
3 years ago
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