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2 years ago

What would be the slope of the line that passses through 4, 5 and 7, 6

1 answer:

7 0

Use the formula:
y2-y1
———
x2-x1

So for this problem it would be:

6-5
——
7-4

= 1
—
3

Hope this helped, if it did I would appreciate it if you would consider giving me brainliest!

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These scatterplots represent the monthly sales and advertising costs of a company. Which trend line is the best fit for the data

Lunna [17]

Answer:

Option D

Step-by-step explanation

I did the assignment

8 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:

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

Percentage of number: 70% of 44

Elena L [17]

0.7 * 44 = 30
30 is seventy percent of forty four

4 0

3 years ago

Read 2 more answers

(50 POINTS - WILL GIVE BRAINLIEST TO MOST HELPFUL ANSWER)

cricket20 [7]

Answer:

Part A:

x + y = 80

x + 20 = y

Part B:

Pam spends 30 minutes practicing math every day.

Part C:

It is not possible for Pam to have spent 60 minutes practicing dance because this means she must have practiced math for 40 minutes (60 - 20 = 40). This would total out to 100 minutes of total practice, not 80 minutes. Therefore, this is impossible.

Step-by-step explanation:

Part A:

"She spends 80 minutes every day practicing dance and math."

x + y = 80

"She dances for 20 minutes longer than she works on math."

x + 20 = y

Part B:

Solve for x:

x + 20 = y

Isolate variable x:

x = y - 20

Plug in this new value for the first equation:

y - 20 + y = 80

Combine like terms:

2y - 20 = 80

Isolate variable y:

2y = 100

y = 50

Plug in the new value of y into any equation:

x + 50 = 80

Isolate variable x:

x = 30

Part C:

x + 20 = y

x + 20 = 60

x = 40

x + y = 80

40 + 60 = 80

100 = 80

Impossible.

8 0

3 years ago

Read 2 more answers

Trigonometry: 1) Work out the bearing from B to C. 2) Calculate the bearing of D from B

kakasveta [241]

Answer:

see below

Step-by-step explanation:

a bearing is the angle in degrees measured clockwise from north.

Triangle ABC is a right triangle

Tan C = opp side / hyp

tan C = AB / CA

tan C = 30/30

tan C = 1

taking the inverse tan

tan ^ -1 tan C = tan ^ -1 ( 1)

C = 45 degrees

This is 90+45 degrees from North

135 degrees from north

Tan B = opp side / hyp

tan B = AD/BA

tan B = 45/30

tan B = 3/2

taking the inverse tan

tan ^ -1 tan B = tan ^ -1 ( 3/2)

D = 56.30993247

Add 180 degrees

180+56.30993247

236.3099325 from north

7 0

3 years ago