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elena-s [515]
2 years ago
15

Analyze the diagram below and complete the instructions that follow.

Mathematics
1 answer:
Damm [24]2 years ago
5 0
D. Angel 6 because vertical is just asking what angle is on the other side of angle 3
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4x-6 + 2x = 18<br> What’s the answer
solmaris [256]

Answer:

x=4

Step-by-step explanation:

4x-6 + 2x = 18

Combine like terms

6x -6 = 18

Add 6 to each side

6x-6+6 =18+6

6x = 24

Divide each side by 6

6x/6 = 24/6

x =4

6 0
3 years ago
The sum of two consecutive integers is greater than 50. Could 25 be one of the numbers?
AlekseyPX

Answer:

Yes. 25 and 26 equal 51, so one of the consecutive numbers could be 25.

Step-by-step explanation:

6 0
3 years ago
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I need an equation for: what number is 70% of 120
kifflom [539]
You'll have to multiply 0.7 by 120.

120*0.7 = n

n is the number that is 70% of 120

120 * 0.7 = 84

Hope this helps :)
8 0
3 years ago
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What is 23 % changed into a fraction
MaRussiya [10]
23% as a fraction would be 23/100
7 0
3 years ago
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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

6 0
1 year ago
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