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3 years ago

What is the slope of the following points (5,7) and (12,2)?

Mathematics

2 answers:

sergey [27]3 years ago

4 0

Answer:

use the slope formula: (y2-y1)/(x2-x1) and then plug in the points given

(2-7)/(12-5)

-5/7

the slope is the last one or "D"

Step-by-step explanation:

stich3 [128]3 years ago

3 0

Answer:

D. -5/7

Step-by-step explanation:

m=y2-y1/x2-x1

m=2-7/12-5

m=-5/7

Hope this helps.

Please mark me brainliest. :)

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2. In what ratio does the point (5, 4) divide the given line joining the points (2,1) and (7,6)

Darya [45]

Answer:

3/2

Step-by-step explanation:

Let it be

C (5;4) , A (2;1) ; B(7;6)

Suppose that C divide the line AB in ratio m/n from point A (AC is m, CB is n)

Use the formula xc=(m*xb+n*xa)/ (m+n)

5=(m*7+2n)/(m+n)

5m+5n= 7m+2n

2m=3n

m/n=3/2

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3 years ago

6 − 3.98 what is the answerrrrr

NARA [144]

Answer:

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3 years ago

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Find the solution r(t)r(t) of the differential equation with the given initial condition: r′(t)=⟨sin4t,sin7t,9t⟩,r(0)=⟨9,8,3⟩

Makovka662 [10]

$\mathbf r'(t)=\langle\sin4t,\sin7t,9t\rangle$
$\implies\mathbf r(t)=\displaystyle\int\mathbf r'(t)\,\mathrm dt$
$\mathbf r(t)=\displaystyle\left\langle\int\sin4t\,\mathrm dt,\int\sin7t\,\mathrm dt,\int9t\,\mathrm dt\right\rangle$
$\mathbf r(t)=\left\langle-\dfrac14\cos4t+C_1,-\dfrac17\cos7t+C_2,-\dfrac92t^2+C_3\right\rangle$

With $\mathbf r(0)=\langle9,8,3\rangle$ , we have

$\langle9,8,3\rangle=\left\langle-\dfrac14+C_1,-\dfrac17+C_2,C_3\right\rangle$
$\implies C_1=\dfrac{37}4,C_2=\dfrac{57}7,C_3=8$
$\implies\mathbf r(t)=\left\langle-\dfrac14\cos4t+\dfrac{37}4,-\dfrac17\cos7t+\dfrac{57}7,-\dfrac92t^2+8\right\rangle$

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4 years ago

Jess wants to buy a car but she can’t decide if she should buy a Honda or a Kia. The Honda costs 16000 and depreciates at an ann

arsen [322]

Answer:

im gonna have my sister answer this one she will be on in a few minutes to answer :D

Step-by-step explanation:

she said give her like 2 minutes and she will be on to answer

5 0

3 years ago

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Need ASAP PLZ 21 POINTS!!!

Dmitriy789 [7]

Answer:

Two of the statements apply:

Second statement: Binh should have graphed the y-intercept of y = x - 3 at (0, - 3)

Last statement: Binh should have found the point of intersection to be (1, - 2).

Explanation:

1. Line 1

$y=-\dfrac{1}{2}x-\dfrac{3}{2}$

2. Line 2

$y=x-3$

3. Binh’s Graph on a coordinate plane, a line with equation y = x - 3 goes through (- 3, 0) and (0, 3).

This is wrong because the point (-3, 0) is not on the line y = x - 3:

x = -3 ⇒ y = -3 - 3 = -6 ⇒ (-3,-6)

The other point, (0, 3) is wrong too

x = 0 ⇒ y = 0 - 3 = - 3 ⇒ (0, -3) This is the y-intercept

4. A line with equation y = -(1/2)x - (3/2) goes through (- 3, 0) and (1, -2).

Check the points:

x = -3 ⇒ y = -(1/2)(-3) - 3/2 = 3/2 -3/2 = 0 ⇒(-3,0) $\checkmark$

x = 1 ⇒ y = -(1/2)(1) - 3/2 = -1/2 -3/2 = -2 ⇒ (1, - 2) $\checkmark$

Both points are correct

5. Binh says the point of intersection is (0, –3). Which statements identify the errors Binh made?

Binh listed the coordinates in the wrong order when describing the point of intersection on his graph?

No, his error was on the procedure, look below

Binh should have graphed the y-intercept of y = x - 3 at (0, - 3)?

Correct!

This was one of the errors, as shown above the y-intercept of the line y = x - 3 is (0, -3) and not (0, 3).

Binh should have graphed the y-intercept of y = x - 3 at (0, 1)?

No. it is (0, 3) as stated above.

Binh should have graphed the y-intercept of y = -(1/2)x -3/2 at (0, -1/2)?

No, this is wrong:

The y-intercept is y = 0 - 3/2 = -3/2 ⇒ (0, -3/2)

Binh should have found the point of intersection to be (1, - 2).

Correct!

Yes, the solution of the equation is (1, -2) and he should have found that intersection point. Proof:

x - 3 = -1/2x - 3/2

x + 1/2x = 3 - 3/2

3/2x = 3/2

x = 1

y = x - 3 ⇒ 1 - 3 = - 2

Thus, the point of intersection is (1, -2).

6 0

4 years ago

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