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3 years ago

6

FindHCF of(726, 255).

1 answer:

Triss [41]3 years ago

3 0

Good evening ,

Answer:

<h2> 3</h2>

Step-by-step explanation:

255 = 3×5×17

726 = 2×3×11^2

Then HCF(255,726) = 3.

:)

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The length of a rectangle is 4 centimeters more than twice its width. If the perimeter of the rectangle is 86 centimeters , find

tensa zangetsu [6.8K]

P = perimeter = 2L + 2W = 86 cm. Also, L = W + 4. Subst. W + 4 for L,

P = 2(W + 4) + 2W = 86 cm. Then 2W + 8 + 2W = 86 cm, and 4W = 78 cm.

Finally solving for W, W = (78 cm)/4, or 19.5 cm.

If W = 19.5 cm, then L = W + 4 cm = 19.5 cm + 4 cm = 23.5 cm

The rectangle's dimensions are 19.5 cm by 23.5 cm.

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2 years ago

Can anyone help me solve this?

Alecsey [184]

Answer:

Step-by-step explanation:

$(\sqrt{z} +\sqrt{22})(\sqrt{z} -\sqrt{22})=35$

Subtract both sides from 35

$(\sqrt{z}+\sqrt{22} (\sqrt{z} -\sqrt{22})-35=35-35$

Simplify:

$(\sqrt{z})^{2} -57=0$

Factor $(\sqrt{z})^{2} -57: (\sqrt{z}+\sqrt{57}) (\sqrt{z}-\sqrt{57})$ =0

$\sqrt{z} +\sqrt{57}=0 or \sqrt{z} -\sqrt{57}=0$

$\sqrt{z} +\sqrt{57} =0: No Solution for Z$

$\sqrt{z} -\sqrt{57} =0: z=57$

Your answer will be z=57

and No solution

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