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Elis [28]
3 years ago
13

The equation y−5=−2(x−3) is in point-slope form. Which shows this equation in slope-intercept form?

Mathematics
1 answer:
MAXImum [283]3 years ago
4 0

y - 5 = -2(x - 3)

Distribute -2.


y - 5 = -2x + 6

Add 5 to both sides.


y = -2x + 11

B) y = -2x + 11

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Find the slop of the line​
Goshia [24]

Answer:

5/4 is your slope.

Step-by-step explanation:

5-0= 5

4-0= 4

slope is chang in y over change in x. Your first point is (0,0) and you second is (4,5).

6 0
3 years ago
Which percent is eguivalent to 2.5?<br><br> 1)2.5%<br> 2)25%<br> 3)250%<br> 4)2,500%
sertanlavr [38]

Answer: 250%

since, 100% = 100/100

250% = 250/100

Step-by-step explanation:

8 0
3 years ago
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Which of the following points are more than 5 units from the point P(−2, −2)? Select all that apply. A A (2, 1) B B (4, −1) C C
netineya [11]

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

<span> |PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}}</span></span>


Using this formula we calculate the distances |PA|, |PB|, |PC|, |PD| and |PE| and compare to 5.


|PA|= \sqrt{ (-2-2)^{2} + (-2-1)^{2}}= \sqrt{16+9}= \sqrt{25}=5

|PB|= \sqrt{ (-2-4)^{2} + (-2+1)^{2}}= \sqrt{36+1}= \sqrt{37} \approx 6

|PC|= \sqrt{ (-2-2)^{2} + (-2+3)^{2}}= \sqrt{16+1}= \sqrt{17}\approx4

|PD|= \sqrt{ (-2+6)^{2} + (-2+6)^{2}}= \sqrt{16+16}= \sqrt{32}\ \textgreater \  \sqrt{25}=5

|PE|= \sqrt{ (-2+5)^{2} + (-2-1)^{2}}= \sqrt{9+9}= \sqrt{16}=4


Answer: B and D





3 0
3 years ago
Choose the equation that satisfies the data in the table.
Zanzabum

Answer:

B. y = \frac{2}{15} X + 2

Step-by-step explanation:

I. Given x = 15 and y = 4

         4 = \frac{2}{15}(15) + 2

         4 = 2+2

         4 = 4 true

II. Give x = 0 and y = 2

         2 = \frac{2}{15}(0) + 2

         2 = 0 + 2

         2 = 2 true

III. Given x = -15 and y = 0

         0 = \frac{2}{15}(-15) + 2

         0 = -2 + 2

        0 = 0 true

Good luck.

6 0
3 years ago
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Anton [14]
Well you have to convert the pounds to ounces
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7 0
3 years ago
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