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3 years ago

9

A number line going from negative 1 and two-thirds to positive 1 and two-thirds in increments of one-third. What is 1 and two-th

irds minus 2 and one-third

2 answers:

serg [7]3 years ago

8 0

Answer:

-2/3

Step-by-step explanation:

Verizon [17]3 years ago

3 0

Answer:

-2/3

that guy is right

Step-by-step explanation:

You might be interested in

<img src="https://tex.z-dn.net/?f=%28y%20-%201%29%28y%20%2B%203%29" id="TexFormula1" title="(y - 1)(y + 3)" alt="(y - 1)(y + 3)"

m_a_m_a [10]

Sense there's only a y, you'd assume y is 1 or 1y.

you'd make the 1 a negative 1 to get rid of the subtraction sign.

you'd add together the 2 ys making it 2y. then add -1 + 3, which would be 2.

so, the answer would be 2y + 2.

7 0

3 years ago

Which of the following shows the factors of 5x2 + 17x + 6? A. (5x + 1)(x + 6) B. (5x + 2)(x + 3) C. (5x + 3)(x + 2) D. (5x + 6)(

zysi [14]

Answer:

B. $(x+3)(5x+2)$

Step-by-step explanation:

The given quadratic trinomial is;

$5x^2+17x+6$

Comparing this to $ax^2+bx+c$ , we have a=5,b=17,c=6.

$ac=5\times6=30$

Two factors of 30 that add up to 17 is 15 and 2.

We split the middle term to obtain;

$5x^2+15x+2x+6$

We factor by grouping;

$5x(x+3)+2(x+3)$

Factor further

$(x+3)(5x+2)$

The correct answer is B.

5 0

3 years ago

Read 2 more answers

6²+(-4)<br> How do u get a answer

kiruha [24]

So 6^2 would be 36 because it is 6 times 6 and then -4 so 36-4=32

7 0

3 years ago

Read 2 more answers

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

Find the amount in a continuously compiunded account for the following condition. Principal, $4000; Annual interest rate, 5.1%;

Ket [755]

A = P*e^(rt)

Here,

A = $4000*e^(0.051*2) = $4000*e^1.02 = $4000(2.773) = $11092.78

Note: This seems very high to me.

7 0

3 years ago