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nikitadnepr [17]
3 years ago
10

Least number by which 1375 is divided to get a perfect sqaure

Mathematics
1 answer:
leonid [27]3 years ago
5 0

Answer:

55

Step-by-step explanation:

1375 = 5 \times 5 \times 5 \times 11 \\  =  {5}^{2}  \times 55 \\

Here 5 is squared and 55 is not.

Hence, the answer is 55

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If f(x) = 2x – 6 and g(x) = 3x + 9, find (f+ g)(x).
Feliz [49]

Answer:

(f + g)(x) = 5x + 3

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>

  • Terms/Coefficients
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Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

f(x) = 2x - 6

g(x) = 3x + 9

(f + g)(x) is f(x) + g(x)

<u>Step 2: Find</u>

  1. Substitute in function values:                                                                          (f + g)(x) = 2x - 6 + 3x + 9
  2. Combine like terms:                                                                                         (f + g)(x) = 5x + 3
4 0
3 years ago
Please help if you can!​
Stolb23 [73]

Answer:

24

Step-by-step explanation:

f(6) = -6

g(5) = -5

Now, we can plug in these values!

-6 - 6(-5) = -6 + 30 = 24

3 0
3 years ago
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You build 7 model airplanes during the summer. At the end of the summer, you have 25 model airplanes. How many model airplanes d
zimovet [89]

Answer:

18

Step-by-step explanation:

You have 25 when you are done, but you made 7, so subtract that from 27.

25-7 equals 18.

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miv72 [106K]

Answer:

23

Step-by-step explanation:

7 0
3 years ago
Multiply: 4x * root(3, 4x ^ 2) * (2 * root(3, 32x ^ 2) - x * root(3, 2x))
Salsk061 [2.6K]

Answer:

32 {x}^{2} \sqrt[3]{ 2x }    -8{x}^{3}

Step-by-step explanation:

We want to

4x \sqrt[3]{4 {x}^{2} } (2 \sqrt[3]{32 {x}^{2} }  - x \sqrt[3]{2x} )

We expand to obtain:

4x \sqrt[3]{4 {x}^{2} }  \times 2 \sqrt[3]{32 {x}^{2} }  -4x \sqrt[3]{4 {x}^{2} } \times  x \sqrt[3]{2x} )

We now simplify

8x \sqrt[3]{4 {x}^{2}  \times 32 {x}^{2} }    -4 {x}^{2}  \sqrt[3]{4 {x}^{2}  \times 2x}

We multiply the radicand

8x \sqrt[3]{64 \times {x}^{3}  \times 2x }    -4 {x}^{2}  \sqrt[3]{8 {x}^{3}}

Or

8x \sqrt[3]{ {(4x)}^{3}  \times 2x }    -4 {x}^{2}  \sqrt[3]{{(2x)}^{3}}

We take cube root to get:

8x  \times 4x\sqrt[3]{ 2x }    -4 {x}^{2}  \times 2x

We multiply out to get:

32 {x}^{2} \sqrt[3]{ 2x }    -8{x}^{3}

6 0
3 years ago
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