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ELEN [110]
3 years ago
14

Sawyer picks a negative number. What is true about the opposite of his number?

Mathematics
2 answers:
BabaBlast [244]3 years ago
6 0
The opposite would be an even number.
Serjik [45]3 years ago
4 0
Sawyer picks a negative number.

The opposite number of a negative number will be it's positive counterpart

the positive number is your answer

hope this helps
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PLEASE HELP ASAP!!!!
mylen [45]
I am pretty sure that the answer is 4/5, hope this helped :)

6 0
2 years ago
Giving brainily if ur correct
frozen [14]

the answer is 60.

all you have to do is multiply all the numbers.

3 × 4 × 5 = 60

first:

multiply two of the numbers together. anything you like.

4 × 3 is 12

and now multiply the left number with the

result.

which is

5 × 12 = 60

8 0
2 years ago
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What's the x-intercept for number 26?
Lunna [17]
It gives function: h = -16t^2 + 40
water is when height is 0 so solve for 0= -16t^2 + 40
subtract 40 from both sides: -40 = -16t^2
divide by -16: -40/-16 = t^2
simplify: 5/2 = t^2
square root both sides: 1.58 = t

the orange hits the river in 1.58 seconds.
6 0
3 years ago
Math question!!!!!!!!!!!!!
BlackZzzverrR [31]

Answer:

the correct answer for that would be identity property

8 0
3 years ago
Read 2 more answers
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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