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3 years ago

Find the square rootV 196

Mathematics

1 answer:

Taya2010 [7]3 years ago

5 0

Answer:

The answer is 14

Step-by-step explanation:

The square root 196

√196 = 14

Thus, The answer is 14

<u>-TheUnknownScientist</u>

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Please help me with this I really need help r-12=25 :r=37

natta225 [31]

Answer:

it's self explanatory 25

Step-by-step explanation:

r -12 = 25

r = 37

37 - 12 = 25

6 0

3 years ago

Solve the following quadratic equation by factoring:<br> <img src="https://tex.z-dn.net/?f=y%5E%7B2%7D" id="TexFormula1" title="

Furkat [3]

Answer:

y = 8 or y = 0

Step-by-step explanation:

Solve for y over the real numbers:

y^2 - 8 y = 0

Factor y from the left hand side:

y (y - 8) = 0

Split into two equations:

y - 8 = 0 or y = 0

Add 8 to both sides:

Answer: y = 8 or y = 0

4 0

2 years ago

notka56 [123]

Answer:

yes razi is correct pateran 3

Step-by-step explanation:

8 0

3 years ago

PLEASE HELP ASAP WILL MARK AS BRAINLYEST

Stolb23 [73]

Answer:

The answer is 40.

Step-by-step explanation:

4x5 for the square part.

2x10 for the rectangle part

add them

5 0

3 years ago

Read 2 more answers

Evaluate lim x→∞ (3x+1)^(4/x), using l'hospital's rule as needed. show all work using proper notation. as you show your work, if

Alborosie

$\displaystyle\lim_{x\to\inty}(3x+1)^{4/x}=\lim_{x\to\infty}e^{\ln(3x+1)^{4/x}}=e^{\lim\limits_{x\to\infty}\ln(3x+1)^{4/x}}$

$\displaystyle\lim_{x\to\infty}\ln(3x+1)^{4/x}=\lim_{x\to\infty}\frac{4\ln(3x+1)}x\stackrel{\mathrm{LHR}}=\lim_{x\to\infty}\frac{4\frac3{3x+1}}1=\lim_{x\to\infty}\frac{12}{3x+1}=0$

$\implies\displaystyle\lim_{x\to\infty}(3x+1)^{4/x}=e^0=1$

4 0

3 years ago

Find the square rootV 196​

Find the square rootV 196