All categories

3 years ago

What numbers could you round to 400,000 if you rounded them to the nearest hundred thousand

Mathematics

2 answers:

elena-s [515]3 years ago

8 0

Answer: the same number rounded to the nearest ten thousand is 350000. Hope it helped! :)

Step-by-step explanation:

Lisa [10]3 years ago

7 0

Answer:

100

Step-by-step explanation:

it is 0 so you would just bump it up by 1

You might be interested in

Which is not a factor of 60? A.4 B.8 C.12 D.15

nordsb [41]

Answer:

Step-by-step explanation:

you could draw a factor tree and list out the factors, however you would not see 8 as an factor

hope this helps!

5 0

2 years ago

Read 2 more answers

Can someone help me with #19

zlopas [31]

Answer:

XZ,X(-10,9)

Y(-4,8)

X2,YZ(14-1)=12

8 0

2 years ago

Brainliest, What is k ?

Ray Of Light [21]

Answer:

k=1

Step-by-step explanation:

Since the coefficient of $x^{3}$ is 1, that is what you put for k

4 0

2 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

2 years ago

I AM GOING TO SCREAM!! I HAVE LITERALLY DRAINED MY POINTS FOR SOMEONE TO ANSWER THIS QUESTION - answer both a and b plz, brainli

sergey [27]

Answer:

a) 12

b) 13

Step-by-step explanation:

f(x) = x^2 + 3

g(x) = 2x - 1

a) f(g(2)) = ?

g(2) = 2(2) - 1

= 4 - 1

= 3

f(g(2)) = f(3) = 3^2 + 3

= 9 + 3

= 12

b) g(f(2)) = ?

f(2) = 2^2 + 3

= 4 + 3

= 7

g(f(2)) = g(7) = 2(7) - 1

= 14 - 1

= 13

7 0

3 years ago