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3 years ago

8

Lucas wants to put soil in his garden shown below. If soil comes in bags that fill 2 square yard each, how many bags should Luca

s buy?

1 answer:

GenaCL600 [577]3 years ago

3 0

does this question have a diagram?if so then plz present it

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Camille bought 3 pounds of nuts for 10.35. What is the unit price per pound?

dusya [7]

3.45. All you have to do is divide 10.35 by 3.

8 0

3 years ago

Read 2 more answers

PLEASE HELP I WILL GIVE BRAINLEST

Ira Lisetskai [31]

The theoretical probability of getting heads or tails is 1/2 the time.

1/2 of 150 = 75

So theoretically you would get 75 heads and 75 tails.

The experiment got 84 heads, which is greater than 75, so it would be higher than the theoretical probability.

The answer is : <span>It is 6% higher than the theoretical probability.</span>

4 0

3 years ago

ANSWER QUICK !!

Nonamiya [84]

507 - 30 = 477

477 / 0.4 = 1192.5

1192.4 / 8 = 149.0625 miles or 150 miles

i don't know if this is right cause the answer seems so absurd

7 0

3 years ago

If a ticket at the movie theater called $7.50 and popcorn cost $3.50 and the drink cost $1.50 and the cab cost $12 for up to fiv

pentagon [3]

This is me cab/ticket,popcorn,drink 7.50+ 3.5+1.50+12.= 24.50

Friends if I take 3

Popcorn,drink,ticket
12.50 ×3= 37.50

So now me plus 3 friends
24.50 +37.5=62.00

8 0

3 years ago

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

3 years ago