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pochemuha
2 years ago
12

Please help with homework. My question is:

le="(x+7)^{4}" alt="(x+7)^{4}" align="absmiddle" class="latex-formula">
Help ASAP! Please! :(
Mathematics
2 answers:
Darya [45]2 years ago
8 0

(x + 7)^{4}

Write 4 as a sum

{(x + 7)}^{2 + 2}

Use \boxed{ \pmb{ {a}^{m + n} =  {a}^{m}   \times  {a}^{n} }} to expand the expression

{(x + 7)}^{2}  \times  {(x + 7)}^{2}

Use \boxed{ \pmb{(a + b)^{2}  =  {a}^{2}  + 2ab +  {b}^{2} }} to expand the expression

( {x}^{2}  + 14x + 49) \times ( {x}^{2}  + 14x + 49)

Multiply the parentheses

{x}^{4}  + 14 {x}^{3}  +  {49x}^{2}  + 14 {x}^{3}  +  {196x}^{2}  + 686x +  {49x}^{2}  + 686x + 2401

Collect like terms

{x}^{4}  +  {28x}^{3}  +  {294x}^{2}  + 1372x + 2401

OLEGan [10]2 years ago
7 0

Answer:

\huge\boxed{ \sf x^4+28x^3+294x^2+1372x+2401}

explanation:

\sf (x+7)^4

\hookrightarrow \sf (x+7) (x+7) (x+7) (x+7)

\hookrightarrow \sf (x^2 +14x+49)(x+7)(x+7)

\hookrightarrow \sf (x^3 +14x^2+49x+7x^2+98x+343)(x+7)

\hookrightarrow \sf (x^3 +21x^2 +147x +343)(x+7)

\hookrightarrow \sf x^4 + 21 x^3 + 147x^2 + 343x + 7x^3 + 147x^2 + 1029x+ 2401

\hookrightarrow \sf x^4+28x^3+294x^2+1372x+2401

note: just keep multiplying and simplifying with patience.

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A) Write 600 as the product of prime factors.<br> Give your answer in index form.
dlinn [17]

Answer:

3 × 5 × 5 × 2 × 2 × 2

Step-by-step explanation:

Make a factor tree.

But let me explain the above to prove it's correct.

3 × 5 = 15

15 × 5 = 75

75 × 2 = 150

150 × 2 = 300

300 × 2 = 600

Therefore 600 as a product of prime factors is: <u>3 × 5 × 5 × 2 × 2 × 2</u>

4 0
3 years ago
Read 2 more answers
Isandro and Yvette each have a calculator. Yvette starts at 100 and subtracts 7 each time. Isandro starts at 0 and adds 3 each t
Talja [164]

Answer: <u>30</u>

Step-by-step explanation:

Yvette- <em>100-7</em>

Isandro- <em>0+3</em>

1.) I'm going to speed up the process and multiply by 5.

Yvette: <em>100-7</em>

7  × 5 = 35

100-35= 65

65-35= 30

Isandro: <em>0+3 </em>

3 × 5 = 15

0+15= 15

15+15=<em> </em>30

2.) If they keep going on this pattern they will have the same number on the 10 round.

7 0
2 years ago
2.26-30 3x+y=6<br>(0,_)<br>(_,0)<br>(3,_)<br>(6,_)<br>(5,_) ​​
bekas [8.4K]

Answer:

Step-by-step explanation:

In ordered pairs (a,b) a is the x value and b is a y value.

if we have 3x+y=6

if x=0, y=6 --> (0,6)

if y=0, x=2 -->(2,0)

if x=3, y=-3--> (3, -3)

if x=6, y= -12 --->(6, -12)

if x=6, y= -9 ---> (5, -9)

7 0
2 years ago
No step by step answers or links
inysia [295]

Answer:

1. k = 15

2. d = 20

3. n = 3

6 0
2 years ago
Which is another way to check the sum of 104+34+228+877?
Alex787 [66]
The answer is digit sum method.

Digit sum method is method used to check the sum of sum numbers. If the sum of all of the digits of numbers is equal to the sum of all of the digits of the total sum, then the arithmetic process was correct.

We need to check the sum of <span>104+34+228+877:
</span>104 + 34 + 228 + 877 = 1243

Let's first summarize the digits of individual numbers:
104 *** 1 + 0 + 4 = 5
34 *** 3 + 4 = 7
228 *** 2 + 2 + 8 = 12 *** 1 + 2 = 3
877 *** 8 + 7 + 7 = 22 *** 2 + 2 = 4

Now, let's sum these sums:
5 + 7 + 3 + 4 = 19 *** 1 + 9 = 10 *** 1 + 0 = <u><em>1</em></u>

Then, let's summarize the digits of the total sum:
1243 *** 1 + 2 + 4 + 3 = 10 *** 1 + 0 = <u><em>1</em></u>

Since the sums of the digits on the both sides of equation is 1, than the arithmetic process was correct and the sum of <span>104 + 34 + 228 + 877 is really 1243.</span>

7 0
3 years ago
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