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3 years ago

I need help please?!!!!

Mathematics

2 answers:

Oksanka [162]3 years ago

8 0

Answer:

69 420

Step-by-step explanation:

ohaa [14]3 years ago

3 0

Answer: Fury

Step-by-step explanation: they have second to most losses.

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(2.3×10^3)+(4.0×10^5)<br>help

DiKsa [7]

4230000 is the answer make me it on this one also

3 0

3 years ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago

When is the best time to use square roots to solve a quadratic? Choose the one BEST

blondinia [14]

Answer: D, When the constants are perfect squares.

Step-by-step explanation:

the “best” method whenever the quadratic equation only contains x2 terms. That implies no presence of any x term being raised to the first power somewhere in the equation.

Hopefully this helps!

8 0

2 years ago

at noon joice drove to the lake at 30 mi. per hour, but she made a long walk back at 4 mi. per hour. how long did she walk ifshe

Kryger [21]

Distance = 30 mph * time
distance = 4 mph * time

distance = 30 mph * t
distance = 4 mph * (17-t)

Since distance is equal then
30 mph * t = 4 mph * (17-t)

30t = -4t + 68

34t = 68

t = 2 hours

We see the return trip time is (17 -t ) which is 15 hours.
The beginning trip time is 2 hours.

Double Check
Beginning trip = 30 miles * 2 = 60 miles
Return trip = 4 miles * 15 = 60 miles.

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B28%20%7D%20%20%2B%20%20%5Csqrt%7B343%7D%20%20%5Cdiv%202%20%5Csqrt%7B63%20%7D%20"

Nadusha1986 [10]

Answer:

Step-by-step explanation:

sqrt(28): sqrt(4*7)

sqrt(4) = 2;

sqrt28)=2*sqrt(7)

sqrt(343): sqrt(7 * 7 * 7) = 7 * sqrt(7)

Note: the rule is if you have 3 equal primes under the root sign, you leave one, you throw one away, and you put one outside the root sign.

2 sqrt(63) = 2 sqrt(3*3*7) The above rule gets modified to throw 1 three away and take the other one outside the root sign.

2sqrt(63) = 2*3 sqrt(7)

Numerator: 2*sqrt(7) + 7sqrt(7) = 9sqrt(7)

9sqrt(7)

======

6 sqrt(7)

3/2

Note without brackets I cannot be certain that I have interpreted this correctly. The division only apply to sqrt(343) / 2 sqrt(63). If this is so please leave a note.

7 0

3 years ago