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Nataly_w [17]
3 years ago
11

Write a slope intercept equation for a line that passes through (-2,5) and (2,-7)

Mathematics
1 answer:
wlad13 [49]3 years ago
7 0

Slope-intercept form: y = mx + b, where m is the slope and b is the y-intercept.

First, find slope using the given coordinates. Formula for slope: y₁ - y₂ / x₁ - x₂.

Our coordinates are (-2, 5) and (2, -7). Plug them in and simplify.

5 - (-7) / -2 - 2

5 + 7 / -4

12/-4

-3

The slope is -3. The equation becomes y = -3x + b.

To find b, plug an (x, y) coordinate on the line in for x and y in the equation and solve. I'll use (-2, 5)

y = -3x + b

5 = -3(-2) + b

5 = 6 + b

-1 = b

The y-intercept is (0, -1). The equation can now be completed!

<h3>Answer:</h3>

y = -3x - 1

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−−→ LO bisects ∠ NLM , LM = 18 , NO = 4 , and LN = 10 . LO → bisects ∠ NLM , LM = 18 , NO = 4 , and LN = 10 . What is the value
Sindrei [870]

Answer:

7.2

Step-by-step explanation:

Given that in triangle LMN, LO is angle bisector of angle L.

LN =10 and LM =18

By angle bisector theorem for triangles we have

LN/LM = NO/MO

Substitute the values for known things and x for MO

We get

10/18 = 4/x

Or cross multiply to get

10x=72

x=7.2

So answer is 7.2

3 0
3 years ago
Find the value of x.<br><br><br> A. 4.3<br> B. 2.7<br> C. 11.6<br> D. 13.9
kakasveta [241]

Answer:

<h2><u>[D] 13.9</u></h2>

Explanation:

  • <em>Pythagorean theorem: a² + b² = c²</em>
  • <em>Solve for hypotenuse (side x) using: c = √a² + b²</em>

12.8² + 5.3² = 191.93

√191.93

= 13.8538803229

<em>Round the answer</em>

13.9

6 0
3 years ago
Find the coordinates of the centroid of triangle RST. SHOW YOUR WORK so I can see if the answer makes sense (no links)
zimovet [89]

Note the coordinates of each point: R(-4, 5), S(5, 1), T(2, -3).

The centroid is the point whose coordinates are the average of the coordinates of R, S, and T.

<em>x</em>-coordinate: (-4 + 5 + 2)/3 = 3/3 = 1

<em>y</em>-coordinate: (5 + 1 - 3)/3 = 3/3 = 1

So the centroid is (1, 1).

3 0
2 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 wasx+\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
Suppose your heart beats 70 times a minute <br>how long would it taake to beat 1 million times
kipiarov [429]
1,000,000/70=approx 142857 mins
142857/60=approx 2381 hours
2381/24=approx 99 days
If you want the answer more precisely then do the same calculations but use the exact value each time
4 0
3 years ago
Read 2 more answers
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