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ser-zykov [4K]
3 years ago
5

Write an algebraic expression for the word phrase 9 less than half a number

Mathematics
1 answer:
Digiron [165]3 years ago
5 0
1/2x -9, you take half of x so multiply 1/2 times x then you subtract the 9 after due to PEMDAS ( parentheses, exponents, multiplication & division, Add & Subtract) you follow the order of operation starting with Parentheses and ending with adding or subtracting
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Someone help me and explain . this is confusing.
pentagon [3]

Answer:

on the image

Step-by-step explanation:

sorry for handwriting lol, hope this helps

7 0
2 years ago
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What is the sum of a number and its additive inverse? A. 0 B. -1 C. 1 D. undefined
Andrej [43]
The answer is A. 0
i hope this helps!
5 0
3 years ago
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PLEASE HELP!!! WILL MARK AS BRAINLISTS!!!
Serjik [45]

The limit is x---->4-

The negative show that x approaches from the left

Now

As x approaches 4 from the left ... Means This number should be less than 4 (<4) but really close to 4.

Let's pick a Number

Say 3.99

Substitute this... You have

3.99/3.99-4

3.99/-0.01

If we choose x to be 3.999

we will have

3.999/-0.001

Notice the pattern... As x approaches 4 from the left... This limit will approach NEGATIVE INFINITY

Why?

As you approach 4 from the left... 3.9,3.99,3.999... You notice that the denominator becomes negative and EXTREMELY SMALL... and when you divide by an extremely small Number..... You'll get a relatively HUGE VALUE(You can try this... Use a calc... Divide any number of choice by a very small number... say.. 0.0000001.... You'll get a huge result

In our case... The denominator is negative... So it Will Approach a very Huge Negative Number

Hence

Answer.. X WILL APPROACH NEGATIVE INFINITY.

Vertical asymptotes are the zeroes of the denominator of a function

The denom. is x-4

Equate to zero to get the asymptote

x-4=0

x=4

Hence... There will be a vertical asymptote at x=4.

Have a great day!

7 0
3 years ago
How many year apart is 384 BC from 1952 AD
julsineya [31]
BC and BCE are Before Common Era
So what you want to do is add 1952 and 384
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Hope this helps
6 0
3 years ago
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For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4
Elina [12.6K]

Answer:

<em>C.</em> (5-\frac{1}{2})^6

Step-by-step explanation:

Given

15(5)^2(-\frac{1}{2})^4

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

Sum = 2 + 4

Sum = 6

Each term of a binomial expansion are always of the form:

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

Where n = the sum above

n = 6

Compare 15(5)^2(-\frac{1}{2})^4 to the above general form of binomial expansion

(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......

Substitute 6 for n

(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

By direct comparison of

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

and

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

We have;

^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4

Further comparison gives

^nC_r = 15

a^{n-r} =(5)^{6-4}

b^r= (-\frac{1}{2})^4

[Solving for a]

By direct comparison of a^{n-r} =(5)^{6-4}

a = 5

n = 6

r = 4

[Solving for b]

By direct comparison of b^r= (-\frac{1}{2})^4

r = 4

b = \frac{-1}{2}

Substitute values for a, b, n and r in

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

Solve for ^6C_4

(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em>(5-\frac{1}{2})^6<em />

3 0
3 years ago
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