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3 years ago

Use the distance formula to find the distance between two points (0.5, 1.3), (-0.4, -1.2)

Mathematics

1 answer:

Alex777 [14]3 years ago

4 0

D=√(0.9²+2.5²)=2.65
The distance between two points is 2.65

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2( 1/5 m− 2/5 )+ 3/5

Liula [17]

Answer:

2m/5-1/5

Step-by-step explanation:

2*m-2/5+3/5

2(m-2)/5+3/5

2m-4/5+3/5

2m/5-1/5

7 0

2 years ago

Mia used 3 different base 10 blocks to model a 3 digit number what is the number?

spin [16.1K]

Answer:

I think its 300

but wouldn't that be like a 2 digit number like 30? scince they are only 10 blocks 10 + 10 + 10 (3×10= 30)

If you dont understand just coment.

5 0

3 years ago

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2.7 is greater than equal to b + 5? Solve

Semenov [28]

The answer: - 2.3 ≥ b ; which does not correspond with any of the answer choices; but most closely corresponds with: "Answer choice: [B]: b > -2.3 ."
_____________
Explanation:
_________________
Assuming we have:
_______________________
2.7 is greater than or equal to "(b + 5)";
_______________________________
We would write:
_________________
→ 2.7 ≥ b + 5 ;
_________________
→ Subtract "5" from EACH side:
_________________
→ 2.7 − 5 ≥ b + 5 − 5

→ - 2.3 ≥ b ; which does not correspond with any of the answer choices; but most closely corresponds with: "Answer choice: [B]: b > -2.3 ."
____________________

7 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

Solve by graphing 2x+4y=8 2x-y=2 point to where they intersect ( , )

miss Akunina [59]

I recommend going to desmos graphing calculator and put those in and it should show

4 0

3 years ago

Read 2 more answers