2 years ago

Can you help me by solving the following question ?

Mathematics

1 answer:

vredina [299]2 years ago

6 0

Answer:

there should be formula at last look carefully and see formula u will find it

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topjm [15]

The Answer is letter a

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3 years ago

Cd + ed = m solve for d

lakkis [162]

Answer:

677

Step-by-step explanation:

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3 years ago

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PLEASE HELP I have been trying to figure this out for a really long time and I want to get the best grade on my test !

Shalnov [3]

Answer:

see below

Step-by-step explanation:

$-4^{2} =-16\\\\-\sqrt{4} =-2\\\\\sqrt{4^{2} } =4\\\\-2\sqrt{4} =-4\\\\(-4)^{2} =16\\\\\sqrt{1/4} =1/2$

so the order

-16

-4

-2

1/2

in root forms and exponent

$-4^{2} \\\\-2\sqrt{4} \\\\-\sqrt{4} \\\\\sqrt{\frac{1}{4} } \\\\\sqrt{4^{2} } \\\\(-4 )^{2}$

8 0

3 years ago

Consider the expansion of (5p + 2q)^6. Determine the coefficients for the terms with the powers of p and q shown.

Step2247 [10]

Answer:

Remember, the expansion of $(x+y)^n$ is $(x+y)^n=\sum_{k=0}^n \binom{n}{k}x^{n-k}y^k$ , where $\binom{n}{k}=\frac{n!}{(n-k)!k!}$ .

Then,

$(5p+2q)^6=\sum_{k=0}^6\binom{6}{k}(5p)^{6-k}(2q)^k=\sum_{k=0}^6\binom{6}{k}5^{6-k}2^k p^{6-k}q^k$

Then, the coefficient of the term $p^{6-k}q^k$ is $\binom{6}{k}5^{6-k}2^k$

a) since 6-k=2, then k=4. So the coefficient of $p^2q^4$ is

$\binom{6}{4}5^{6-4}2^4=15*5^2*2^4=15*25*16=6000$

b) since 6-k=5, then k=1. So, the coefficient of $p^5q$ is

$\binom{6}{1}5^{6-1}2^1=6*5^5*2=37500$

c) since 6-k=3, then k=3. So, the coefficient of $p^3q^3$ is

$\binom{6}{3}5^{6-3}2^3=20*5^3*8=20000$

6 0

3 years ago

What should the following equation be multiplied by to eliminate the decimals? -200k – 3.01 = 3.3

True [87]

Answer: 100

-200k - 3.01 = 3.3 | · 100

-200k · 100 - 3.01 · 100 = 3.3 · 100

-20,000k - 301 = 330

5 0

3 years ago