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3 years ago

use the slope formula to calculate the slope of the line that contains the two points (-4,0) and (5,2)

Mathematics

1 answer:

Solnce55 [7]3 years ago

3 0

Answer:

$\frac{2}{9} x$

Step-by-step explanation:

To find slope using two points you need to find the difference in x and y. You use $\frac{y_{2}-y_{1}}{x_{2}-x_{1} }$

So, 2-0=2 and 5-(-4)=9. Therefore the slope is 2/9x

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Wendy throws a dart at this square-shaped target:

Ostrovityanka [42]

Answer:

Hello! I hope I am correct! :)

Step-by-step explanation:

Let’s first calculate the black circle & the white space.

Since there is more white space than the black circle, we already know that the white space will have more probability.

These are the steps you need to do, in order to so love this problem:

1. Find the area of both, the white space and the black circle.

2. Divide both of them by the area of the square.

3. Do all these steps to find the probability/hitting.

Part A:

Black circle: ( π *1^2) / ( π 5^2) = 1/25 = .04 or 3.14% chance.

So we can tell it’s close to zero.

Part B: 1 - .04 = 0.96 or 97%

This is close to one.

(Remember I am not saying the exact, it’s a estimate)

Brainliest would be appreciated!

Hope this helps! :)

By; BrainlyAnime

8 0

3 years ago

Simplify (12a5−6a−10a3)−(10a−2a5−14a4) . Write the answer in standard form

almond37 [142]

Answer:

$=14a^5+14a^4-10a^3-16a$

Step-by-step explanation:

$\left(12a^5-6a-10a^3\right)-\left(10a-2a^5-14a^4\right)\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=12a^5-6a-10a^3-\left(10a-2a^5-14a^4\right)\\-\left(10a-2a^5-14a^4\right):\quad -10a+2a^5+14a^4\\-\left(10a-2a^5-14a^4\right)\\\mathrm{Distribute\:parentheses}\\=-\left(10a\right)-\left(-2a^5\right)-\left(-14a^4\right)\\Apply\:minus-plus\:rules\\-\left(-a\right)=a,\:\:\:-\left(a\right)=-a\\=-10a+2a^5+14a^4\\=12a^5-6a-10a^3-10a+2a^5+14a^4$

$\mathrm{Simplify}\:12a^5-6a-10a^3-10a+2a^5+14a^4:\quad 14a^5+14a^4-10a^3-16a12a^5-6a-10a^3-10a+2a^5+14a^4\\Group\:like\:terms\\=12a^5+2a^5+14a^4-10a^3-6a-10a\\\mathrm{Add\:similar\:elements:}\:12a^5+2a^5=14a^5\\=14a^5+14a^4-10a^3-6a-10a\\\mathrm{Add\:similar\:elements:}\:-6a-10a=-16a\\=14a^5+14a^4-10a^3-16a$

4 0

3 years ago

Round off the following numbers to two decimal places <br> 1.2.1) 0,77677<br> 1.2.2) 34,2784682

andre [41]

1) 0.78
2) 34.28
This is because the number in the 3rd decimal place for both numbers is 5 or above so it is rounded up rather than down.

8 0

3 years ago

Which equation represents the line that passes through (-6, 7) and (-3, 6)?

Degger [83]

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:

$y = mx + b$

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis

We have the following points through which the line passes:

$(x_ {1}, y_ {1}): (- 6,7)\\(x_ {2}, y_ {2}): (- 3,6)$

We find the slope of the line:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {6-7} {- 3 - (- 6)} = \frac {-1} {-3 + 6} = \frac {-1} {3} = -\frac {1} {3}$

Thus, the equation of the line is of the form:

$y = - \frac {1} {3} x + b$

We substitute one of the points and find b:

$6 = - \frac {1} {3} (- 3) + b\\6 = 1 + b\\b = 5$

Finally, the equation is:

$y = - \frac {1} {3} x + 5$

Answer:

$y = - \frac {1} {3} x + 5$

8 0

3 years ago

Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the

Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees was $x+\frac{1}{4} x=\frac{5}{4} x$ , and for the next three years we have that

Start End

Second year $\frac{5}{4}x$ -------------- $\frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x$

Third year $(\frac{5}{4} )^{2}x$ ------------- $(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x$

Fourth year $(\frac{5}{4})^{3}x$ -------------- $(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.$

So the formula to calculate the number of trees in the fourth year is

$(\frac{5}{4} )^{4} x,$ we know that all of the trees thrived and there were 6250 at the end of 4 year period, then

$6250=(\frac{5}{4} )^{4}x$ ⇒ $x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.$

Therefore the number of trees at the begging of the 4-year period was 2560.

7 0

3 years ago

use the slope formula to calculate the slope of the line that contains the two points (-4,0) and (5,2)​

use the slope formula to calculate the slope of the line that contains the two points (-4,0) and (5,2)