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3 years ago

how many of the cookies had fewer than 6 chocolate chips in them? 5,6,3,9,6. What is the average amount of chocolate chips?

1 answer:

4 0

2 cookies had less than 6 if you are saying those numbers are cookies with the number of chocolate chips. And the average amount of chocolate chips would be 5.8

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if α,β are the roots of the equation x^{2}-3x+2=0 form an equation whose roots are (α+β)² & (α-β)²

GuDViN [60]

Solution:-

Let's find roots of x² - 3x + 2

=> x² - 3x + 2 = 0.

=> x² - x - 2x + 2 = 0.

=> x ( x - 1 ) -2 ( x - 1) = 0.

=> ( x - 1 ) ( x - 2 ) = 0.

=> x = 2, 1.

Now

( a + ß )² = (2+1)²=3²=9
( a - ß )² = ( 2 - 1 )² = 1² = 1.

So , equⁿ would be ,

=> x² - ( 9 + 1)x + 9×1=0.

=> x² - 10x + 9=0.

6 0

2 years ago

Root 3 cosec140° - sec140°=4 prove that

Lorico [155]

Answer:

Step-by-step explanation:

We are to show that $\sqrt{3} cosec140^{0} - sec140^{0} = 4\\$

Proof:

From trigonometry identity;

$cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}$

$\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\$

From trigonometry, 2sinAcosA = Sin2A

$= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\= \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2 \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}$

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

$\frac{4(sin(420-140)) }{sin280}$

= $\frac{4sin280}{sin280}\\ = 4 \ Proved!$

3 0

2 years ago

Can someone help me I need the answers to 1-4 ty!!

lubasha [3.4K]

1. x=-3 y=4 2. x=5/2 y=15/2 3. x=9/5 y=-36/5 4. x=-2 y=1

8 0

2 years ago

What is sin (30 deg)?

pav-90 [236]

Answer:

1/2

Step-by-step explanation:

Other answers ; 0.5,2^-1

8 0

2 years ago

What is the product?

Sophie [7]

Answer:

Step-by-step explanation:

In exponent multiplication, if base are same just add the powers.

$(-6a^{3}b+2ab^{2})(5a^{2}-2ab^{2}-b)=(-6a^{3}b)*(5a^{2}-2ab^{2}-b)+2ab^{2}*(5a^{2}-2ab^{2}-b)\\\\$

$=(-6a^{3}b)*5a^{2}-(-6a^{3}b)*2ab^{2}-(-6a^{3}b)*b+ 2ab^{2}*5a^{2}-2ab^{2}*2ab^{2}-2ab^{2}*b\\\\=-30a^{3+2}b+12a^{3+1}b^{1+2}+6a^{3}b^{1+1}+10a^{1+2}b^{2}-4a^{1+1}b^{2+2}-2ab^{2+1}\\\\=-30a^{5}b+12a^{4}b^{3}+6a^{3}b^{2}+10a^{3}b^{2}-4a^{2}b^{4}-2ab^{3}\\$

6 0

2 years ago