Math, asked by testsix5118, 9 months ago

56 , 64 and 104 find the HCF using prime factorization

Answers

Answered by Anonamyms

1

✌️Hey mate ✌️

Here's the answer :

$</em></strong><strong><em> HCF</em></strong><strong><em>(56</em></strong><strong><em>,</em></strong><strong><em>64</em></strong><strong><em>,</em></strong><strong><em>104) = 2 \times 2 \times 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 8$

Kindly refer to the Attachment

Kindly rate and like

Mark as Brainliest

▄▀▄▀▄▀ ғᴏʟʟᴏᴡ ▄▀▄▀▄▀

Attachments:

Previous Question

Next Question

Similar questions

Math, 4 months ago

Playing With Whole Numbers 71 7. Find the smallest number which when divided by 21, 28, 36 and 45 leaves a remainder 9 in each case. Find the...

Math, 4 months ago

if 2 sin 2 theta = root3,then theta = ...

History, 4 months ago

surdas wrote sursugar surasaravali and........

English, 9 months ago

Students have completed their work in passive voice...

Science, 9 months ago

The Earth’s history has had a significant effect on the characteristics of its organisms and biomes. How was the Carboniferous period different from the Devonian period?

Chemistry, 1 year ago

The species which by definition has ZERO standard molar enthalpy of formation at 298 K is (a) Br₂ (g) (b) Cl₂ (g) (c) H₂O (g) (d) CH₄ (g)...

Chemistry, 1 year ago

On the basis of the following thermochemical data : $[\Delta_fH^0 (H^+_{(aq)}= 0)]$ $H_2O(l) \rightarrow H^+(aq) + OH^-(aq)$ ; $\Delta H$ = 57.32kJ $H_2(g) + \frac{1}{2}O_2(g) \longrightarrow H_2O(l)$ ; $\Delta H$ = - 286.20kJ The value of enthalpy of formation of OH⁻ ion at 25° C is: (a) –228.88 kJ (b) +228.88 kJ (c) –343.52 kJ (d) –22.88 kJ...

Chemistry, 1 year ago

Consider the reaction $4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g),$ $\Delta_rH$ = – 111 kJ. If N2O5 (s) is formed instead of N₂O₅ (g) in the above reaction, the $\Delta_rH$ value will be :(given, $\Delta H$ of sublimation for N₂O₅ is –54 kJ mol⁻¹) (a) + 54 kJ (b) + 219 kJ (c) – 219 J (d) – 165 kJ...