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3 years ago

What is the slope of the line?

Mathematics

2 answers:

postnew [5]3 years ago

4 0

Answer:

its 2

Step-by-step explanation:

rise/run. it goes up one and over .5 each time, so 1÷0.5=2.

pashok25 [27]3 years ago

4 0

Answer:

1/2

Step-by-step explanation:

You need to find the point that is connected.

Hope this helps.

You might be interested in

7/9 + 5/12 with lcm pls help and give explanation pls

Viefleur [7K]

Answer:

1. 43/36

2. 1/2

Step-by-step explanation:

1. 7/9 + 5/12

= LCM of 9 and 12 is 36

= 28/36 + 15/12

= (28 + 15) / 36

= 43/36

2. 21/24 - 3/8

= LCM of 24 and 8 is 24

= 21/24 - 9/24

= (21 - 9) / 24

= 12/24

= 1/2

3 0

2 years ago

Read 2 more answers

Barbie is thinking of a number 20 less than one-third of the number is 72 find barbie's number

ExtremeBDS [4]

Let x = Barbie's number
1/3 x -20 = 72
1/3 x = 92
x = 92*3=276

4 0

3 years ago

John has a piece of clay that weighs 12 1/2 pounds. He divides the clay into 5 equal pieces. How much does each piece weigh?

djyliett [7]

Answer:

2.44

Step-by-step explanation:

12 1/2 / 5 =2.44

3 0

3 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:

C. $(5-\frac{1}{2})^6$

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]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

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

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 $^6C_4$

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

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

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

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

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

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

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

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

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago

10+5-7=3y -10x+5x+5=-15 3x+2x-4x+10 pre algebra 8th grade math

jasenka [17]

Answer:

1) y=8/3=2 2/3 2) x=4 3) x+10

Step-by-step explanation:

1) 10+5-7=3y

Calculate the sum or difference.

8=3y

Swap the sides of the equation.

3y=8

Divide both sides by 3.

y=8/3=2 2/3

2) -10x+5x+5=-15

Collect like terms.

-5x+5=-15

Move the constant to the right-hand side and change the sign.

-5x=-15-5

Calculate the difference.

-5x=-20

Divide both sides by -5.

x=4

3) 3x+2x-4x+10

Collect like terms.

(3+2-4)x+10

Calculate the sum or difference.

1x+10

When the term has a coefficient of 1, it doesn't have to be written.

x+10

Hope this helps :)

5 0

2 years ago