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lutik1710 [3]
3 years ago
8

How to solve 52/48 in the form of hcf​

Mathematics
1 answer:
algol133 years ago
6 0

HCF is 4

Step-by-step explanation:

factors of 52 are: 2x2x13

factors of 48 are:2x2x2x2x3

common factors of 52%48 are: 2

hence, = 2x2

=4

hope it helps

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2(3x+2)=2x-1+x solve
alina1380 [7]
First multiply and dstribute
a(b+c)=ab+ac
2(3x+2)=6x+4
6x+4=2x-1+x
add like terms
6x+4=3x-1
subtract 3x from both sides
3x+4=-1
subtract 4 from both sides
3x=-5
divide bothe sides by 3
x=-5/3
7 0
3 years ago
41 5<br> ---- = ----- x=?<br> x 2
Stells [14]
\frac{41}{x}=\frac{5}{2}\\&#10;5x=82\\&#10;x=\frac{82}{5}
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A recipe for cookies calls for 1/3 of a cup of sugar per batch. Elena used 5 1/3 cups of sugar to make multiple batch of cookies
shusha [124]

Answer:

she can make 16 batches

Step-by-step explanation: simply multiply 5 by 3 and add 1(for the fraction)

6 0
3 years ago
Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8​
Ivanshal [37]

It looks like the differential equation is

\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}

Factorize the right side by grouping.

xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)

xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)

Now we can separate variables as

\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx

Integrate both sides.

\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx

\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx

\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}

You could go on to solve for y explicitly as a function of x, but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0
2 years ago
Find f(-2) for f(x) = 2(4)^x<br><br> O A.1/16<br> O B.1/8<br> O C. -16<br> OD. - 32
scoray [572]

Answer:

B. 1/8

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Function Notation

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle f(x) = 2(4)^x\\f(-2) \ is \ x = -2<u />

<u />

<u>Step 2: Evaluate</u>

  1. Substitute in <em>x</em>:                    \displaystyle f(-2) = 2(4)^{-2}
  2. Exponents:                          \displaystyle f(-2) = 2(\frac{1}{16})
  3. Multiply:                               \displaystyle f(-2) = \frac{1}{8}
5 0
3 years ago
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