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3 years ago

12

2 1/2 + 2 1/3 + 3 2/3

1 answer:

IrinaVladis [17]3 years ago

6 0

Answer:

8 1/2

Step-by-step explanation:

2 1/2 + 2 1/3 + 3 2/3

= 2 3/6 + 2 2/6 + 3 4/6

= 7 9/6

= 8 3/6

= 8 1/2

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How do u write any number to the zeroth power

SSSSS [86.1K]

Answer:

0 as an exponent would look something like 6^0 and anything to the power of 0 is always 1

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2 years ago

Hall and Mindy are playing a guessing game. Hall tells Mindy: ""The difference between 17 and the square root of my mystery numb

cestrela7 [59]

Answer:

The possible number that Hall could be thinking of is 144.

Step-by-step explanation:

Difference between 17 and the square root of my mystery number

This is represented by:

$17 - \sqrt{x}$

is 5

Then:

$17 - \sqrt{x} = 5$

What are two possible numbers that Hall could be thinking of?

We have to solve the equation, for x. So

$17 - \sqrt{x} = 5$

$\sqrt{x} = 12$

$(\sqrt{x})^2 = (12)^2$

$x = 144$

The possible number that Hall could be thinking of is 144.

5 0

3 years ago

Which law would you use to simplify the expression (x^4)^9

drek231 [11]

Power of a power is the law<span />

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2 years ago

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

2 years ago

50 PTS ANSWER ALL <3333333

11Alexandr11 [23.1K]

QUESTION 33

The length of the legs of the right triangle are given as,

6 centimeters and 8 centimeters.

The length of the hypotenuse can be found using the Pythagoras Theorem.

${h}^{2} = {6}^{2} + {8}^{2}$

${h}^{2} = 36+ 64$

${h}^{2} = 100$

$h = \sqrt{100}$

$h = 10cm$

Answer: C

QUESTION 34

The triangle has a hypotenuse of length, 55 inches and a leg of 33 inches.

The length of the other leg can be found using the Pythagoras Theorem,

${l}^{2} + {33}^{2} = {55}^{2}$

${l}^{2} = {55}^{2} - {33}^{2}$

${l}^{2} = 1936$

$l = \sqrt{1936}$

$l = 44cm$

Answer:B

QUESTION 35.

We want to find the distance between,

(2,-1) and (-1,3).

Recall the distance formula,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the values to get,

$d=\sqrt{( - 1-2)^2+(3- - 1)^2}$

$d=\sqrt{( - 3)^2+(4)^2}$

$d=\sqrt{9+16}$

$d=\sqrt{25}$

$d = 5$

Answer: 5 units.

QUESTION 36

We want to find the distance between,

(2,2) and (-3,-3).

We use the distance formula again,

$d=\sqrt{( - 3-2)^2+( - 3- 2)^2}$

$d=\sqrt{( - 5)^2+( - 5)^2}$

$d=\sqrt{25+25}$

$d=\sqrt{50}$

$d=5\sqrt{2}$

Answer: D

8 0

2 years ago

Read 2 more answers