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marissa [1.9K]
2 years ago
10

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
8 0
3 years ago
Cd + ed = m solve for d
lakkis [162]

Answer:

677

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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

4

16

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
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