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3 years ago

10

On graph paper, draw diagrams of a+a+a+a and 4a when a is 1, 2, and 3. What do you notice?

2 answers:

Bess [88]3 years ago

7 0

We notice that, 4a = a + a + a + a

olasank [31]3 years ago

3 0

4a is added 4x which means a+a+a+a

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A right triangle has a side with length 12 in and a hypotenuse with length 20 in. find the length of the missing leg. (round to

OLEGan [10]

Let the unknown side be "z" inches
Through pythagoras:
${12}^{2} + {z}^{2} = {20}^{2}$
Because the square of the two sides add up to the square of the hypotenuse in a right triangle.

This means that
${z }^{2} = {20}^{2} - {12}^{2}$
${z}^{2} = 400 - 144 = 256$
$z = \sqrt{256} = 16$
So the missing side is 16 inches

Hope this helped

6 0

3 years ago

Indicate a general rule for the nth term of this sequence. -6b, -3b, 0b, 3b, 6b. . .

GREYUIT [131]

Answer:

a(n) = -16b + (n - 1)(3b)

Step-by-step explanation:

First term is -6b.

Common difference is 3b; each new term is equal to the previous one, plus 3b.

Formula for this arithmetic sequence is

a(n) = -16b + (n - 1)(3b)

4 0

3 years ago

I wanna cry Joseph pls answer this

Xelga [282]

I ain't Joseph but what's wrong?

5 0

2 years ago

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

4 0

3 years ago

Need help ASAP!!!!!!!!!!!

8090 [49]

Answer:

Step-by-step explanation:

5/6 + 1/8 + 3/4 = 1 17/24

Option A is the correct answer

8 0

3 years ago