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2 years ago

All numbers more than 2 units from 6

Mathematics

1 answer:

IRINA_888 [86]2 years ago

5 0

All numbers more than 2 units from 6 are greater than or equal to 8

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If some one has a circle that is 3 radius what is the diameter

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Answer:

The diameter is 6

Step-by-step explanation:

the radius is just half of the diameter so 2x3= 6

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2 years ago

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2 years ago

The measure of the supplement of an angle is three times the measure of the

gtnhenbr [62]

<h2>Answer:</h2>

<h3><em>x=45degrees</em></h3>

<h2>Step-by-step explanation:</h2>

Let the angle to be solved be x

Let the supplement/compliment by y

x+y=90 Complimentary angles add up to 90 degrees.

x+3y=180 Supplementary angles add up to 180 degrees, the other angle is thrice the other compliment.

Evaluating this as a system:

x+y=90 Isolate x:

x=90−y Input into the other equation:

(90−y)+3y=180 Combine like terms, isolate y and its coefficients:

2y=90 Isolate y

y=45 Input into the first equation:

x+45=90 Isolate x:

x=45degrees

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2 years ago

What is cos 30°? 60° 1 90° 30° V3 O A. v3 O B. 1 O c.

siniylev [52]

Answer:

D. 1/2

Step-by-step explanation:

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2 years ago

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Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.

Mice21 [21]

Answer:

<em>LCM</em> = $8y^{4}(y+ 10)^{2}(y + 8)$

Step-by-step explanation:

Making factors of $8y^{6}+ 144y^{5}+ 640y^{4}$

Taking $8y^{4}$ common:

$\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)$

Using <em>factorization</em> method:

$\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times 4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)$

Now, Making factors of $2y^{4} + 40y^{3} + 200y^{2}$

Taking $2y^{2}$ common:

$\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)$

Using <em>factorization</em> method:

$\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10) ............ (2)$

The underlined parts show the Highest Common Factor(HCF).

i.e. <em>HCF</em> is $2y^{2} (y+ 10)$ .

We know the relation between <em>LCM, HCF</em> of the two numbers <em>'p' , 'q'</em> and the <em>numbers</em> themselves as:

$HCF \times LCM = p \times q$

Using equations <em>(1)</em> and <em>(2)</em>: $\Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)$

Hence, <em>LCM</em> = $8y^{4}(y+ 10)^{2}(y + 8)$

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2 years ago