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3 years ago

11

* URGENT* help plz plz plz plz plz plz plz I SPENT FOREVERRRRRRRRRRR

1 answer:

kiruha [24]3 years ago

7 0

answer:

4(2+x) and 8 + 4x

since she memorized those scales for EACH of those four lessons, the answer would be that.

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Is -10x^8 a monomial.?

tensa zangetsu [6.8K]

Yep! because there is no +,-, ect. sign. in total it is just one number

7 0

3 years ago

Read 2 more answers

Adam will be 50 years old in eleven years. How old is he now?

user100 [1]

Answer:

39 years old

Step-by-step explanation:

50 - 11 = 39 years old

6 0

3 years ago

What is the 4th term of the expanded binomial (2x – 3y)^6

san4es73 [151]

Answer:

The 4th term of the expanded binomial is $-4320x^3y^3$

Step-by-step explanation:

Considering:

$(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k}y^k$

$(2x-3y)^6 = \sum_{k=0}^{6} \binom{6}{k} (2x)^{6-k}(-3y)^k$

Now, you gotta calculate for every value of $k$

$(2x-3y)^6 = \binom{6}{0} (2x)^{6-0}(-3y)^0 + \binom{6}{1} (2x)^{6-1}(-3y)^1 + \binom{6}{2} (2x)^{6-2}(-3y)^2 + \\$

$\binom{6}{3} (2x)^{6-3}(-3y)^3 + \binom{6}{4} (2x)^{6-4}(-3y)^4 + \binom{6}{5} (2x)^{6-5}(-3y)^5 + \binom{6}{6} (2x)^{6-6}(-3y)^6$

I will not write every product, but just solve following the steps:

For $k=0$

$\binom{6}{0} (2x)^{6-0}(-3y)^0$

$\frac{6!}{(6-0)!(0!)} (2x)^{6-0}(-3y)^0$

$\frac{6!}{6!} \left(2x\right)^{6-0}\cdot 1$

$1\cdot \:1\cdot \left(2x\right)^{6-0}$

$2^6x^6$

$64x^6$

$(2x-3y)^6=64x^6-576x^5y+2160x^4y^2-4320x^3y^3+4860x^2y^4-2916xy^5+729y^6$

8 0

3 years ago

Fernando followed two diagonal paths, Paths 1 and 2, to get from his house, F, to a neighborhood corner store, C, as shown below

aleksandrvk [35]

Answer:

42 meters

Step-by-step explanation:

Given

See attachment for complete question

Required

Calculate distance FC

FC is calculated as:

$FC = Path\ 1 + Path\ 2$

Where

$Path\ 1 = FB$

$Path\ 2 = BC$

Considering $\triangle FEB$ , we have:

$FB^2 = FE^2 + BE^2$ --- Pythagoras theorem

Where

$FE = FD - BA$

$FE = 22m - 15m$

$FE = 7m$

and

$BE = CD - CA$

$BE = 32m - 8m$

$BE = 24m$

So:

$FB^2 = FE^2 + BE^2$

$FB^2 = 7^2 +24^2$

$FB^2 = 49 +576$

$FB^2 = 625$

Take square roots of both sides

$FB = 25$

So:

$Path\ 1 = 25$

Considering $\triangle BAC$ , we have:

$BC^2 = BA^2 + AC^2$ --- Pythagoras theorem

Where:

$BA = 15$ and $AC = 8$

So, we have:

$BC^2 = 15^2 + 8^2$

$BC^2 = 225 + 64$

$BC^2 = 289$

Take square roots of both sides

$BC = 17$

So;

$Path\ 2= 17$

Recall that:

$FC = Path\ 1 + Path\ 2$

$FC = 25 + 17$

$FC = 42$

6 0

3 years ago

6834/17 please show your work

Sunny_sXe [5.5K]

Answer:

6834 / 17 = 402

Step-by-step explanation:

• sorry i d k, I just got the ans on calculator

3 0

2 years ago