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2 years ago

What is the answer to this question ? Pls help!!

Mathematics

2 answers:

sammy [17]2 years ago

6 0

The answer is B when you subtract every like term

pickupchik [31]2 years ago

5 0

Attached a solution and showed work.

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Is 63 greater than 48.1 or less than PLEASE HELP QUICKLY AS POSSIBLE THANK YOU :)

timofeeve [1]

Answer:

63 is greater than 48.1 ...

63 > 48.1

Step-by-step explanation:

63 is greater than 48 remember. So you need to remove the [.1] and focus on the 63 and 48.

<h3>Hope it helps!!</h3><h3>Please mark me as the brainliest!!!</h3>

Thanks!!!!❤❣❤

5 0

3 years ago

Read 2 more answers

What is the value of x? (2x-54) (16)°

vagabundo [1.1K]

Answer:

Step-by-step explanation:

7 0

2 years ago

A projectile is launched into the air. The function h(t) = –16t2 + 32t + 128 gives the height, h, in feet, of the projectile t s

Zinaida [17]

Answer:

t = 4 seconds

Step-by-step explanation:

The height of the projectile after it is launched is given by the function :

$h(t)=-16t^2+32t+128$

t is time in seconds

We need to find after how many seconds will the projectile land back on the ground. When it land, h(t)=0

So,

$-16t^2+32t+128=0$

The above is a quadratic equation. It can be solved by the formula as follows :

$t=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

Here, a = -16, b = 32 and c = 128

$t=\dfrac{-32\pm \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=\dfrac{-32+ \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}, \dfrac{-32\- \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=-2\ s\ \text{and}\ 4\ s$

Neglecting negative value, the projectile will land after 4 seconds.

4 0

3 years ago

NEED HELP ASAP!! WILL MARK BRAINLIEST!!!!

dmitriy555 [2]

A = 11/5, let me know if you need me to tell you how i got it (show my work)

4 0

3 years ago

The nth term of a sequence is 3n²-1

Anarel [89]

Answer: The number is 26.

Step-by-step explanation:

We know that:

The nth term of a sequence is 3n²-1

The nth term of a different sequence is 30–n²

We want to find a number that belongs to both sequences (it is not necessarily for the same value of n) then we can use n in one term (first one), and m in the other (second one), such that n and m must be integer numbers.

we get:

3n²- 1 = 30–m²

Notice that as n increases, the terms of the first sequence also increase.

And as n increases, the terms of the second sequence decrease.

One way to solve this, is to give different values to m (m = 1, m = 2, etc) and see if we can find an integer value for n.

if m = 1, then:

3n²- 1 = 30–1²

3n²- 1 = 29

3n² = 30

n² = 30/3 = 10

n² = 10

There is no integer n such that n² = 10

now let's try with m = 2, then:

3n²- 1 = 30–2² = 30 - 4

3n²- 1 = 26

3n² = 26 + 1 = 27

n² = 27/3 = 9

n² = 9

n = √9 = 3

So here we have m = 2, and n = 3, both integers as we wanted, so we just found the term that belongs to both sequences.

the number is:

3*(3)² - 1 = 26

30 - 2² = 26

The number that belongs to both sequences is 26.

6 0

2 years ago