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2 years ago

Are the expressions bh/2 and (1/2)bh equivalent expressions?

Mathematics

1 answer:

andreev551 [17]2 years ago

4 0

Bh/2 = (1/2)bh, so <span>bh/2 and (1/2)bh equivalent expressions</span>

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Help me asap! I will give you marks

Law Incorporation [45]

Recall the binomial theorem.

$(a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k$

1. The binomial expansion of $\left(1+\frac x3\right)^7$ is

$\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}$

Observe that

$k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x$

$k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2$

When we multiply these by $8-9x$ ,

• $8$ and $\frac73 x^2$ combine to make $\frac{56}3 x^2$

• $-9x$ and $\frac73 x$ combine to make $-\frac{63}3 x^2 = -21x^2$

and the sum of these terms is

$\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}$

2. The binomial expansion is

$\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k$

We get the $a^6b^2$ term when $k=2$ :

$k=2 \implies \dbinom 82 2^{8-2\cdot2} a^{8-2} b^2 = 28 \cdot2^4 a^6 b^2 = \boxed{448} \, a^6b^2$

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1 year ago

Hello can someone send me some documment for geometry where are the signs explained(Like )

Ket [755]

Answer:

https://revisionmaths.com/gcse-maths-revision/shape-and-space/vectors

i find this helpful- hope u do too (can u mark my answer as brainliest?)

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It would go to quadrent 3

Step-by-step explanation:

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Assoli18 [71]

Answer:

Step-by-step explanation:first:4+8 then 12*2 after:24+4

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Is a location that has no size, represented by a small dot.

monitta

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