2 years ago

Find the highest common factor(HCF) of 18 and 24

Mathematics

1 answer:

prisoha [69]2 years ago

6 0

18 - 2,3,6,8
24 - 2,4,6,8,12

HCF = 8

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cricket20 [7]

Sorry, I won't understand

6 0

3 years ago

Jake leaned a 12-foot ladder against his house. If the Angle formed by the ladder and the ground is 68, how far from the base of

zhuklara [117]

Answer:

$4.50\ ft$

Step-by-step explanation:

see the attached figure to better understand the problem

In the right triangle ABC

$cos(A)=\frac{AC}{AB}$ -----> the cosine is the adjacent side to angle A divided by the hypotenuse

substitute the values and solve for x

$cos(68\°)=\frac{x}{12}$

$x=12*cos(68\°)=4.50\ ft$

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3 years ago

3, 1-을<br>6. +<br>9. 1-(급+음<br>12. 37 -27

umka21 [38]

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3 years ago

If you deposit $50 in checks in your account with $100 cash amount, what is your total deposit?

Free_Kalibri [48]

What do you mean? You deposited $50 didn’t you? HUuH

3 0

2 years ago

Evaluate using <br> Definite integrals

swat32

Since $[0,4]=[0,1]\cup(1,4]$ , we can rewrite the integral as

$\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt$

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

$\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt$

Both integrals are quite immediate: you only need to use the power rule

$\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}$

to get

$\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4$

Now we only need to evaluate the antiderivatives:

$\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15$

So, the final answer is 15.

4 0

3 years ago