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2 years ago

Factorise 10xy-12+15x-8y

Mathematics

1 answer:

GaryK [48]2 years ago

4 0

Answer: 10xy-12+15x-8y=(5x-4)*(2y+3).

Step-by-step explanation:

$10xy-12+15x-8y=(10xy-8y)+(15x-12)=\\=2y*(5x-4)+3*(5x-4)=(5x-4)*(2y+3).$

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Alex17521 [72]

The answer is C. 16 since the 3 first surveys medians are 15,16,16. The median of those 3 numbers is 16. So the answer would be C. 16

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3 years ago

I NEED HELP what is the slope intercept form of y+6=4/5(x+3)

Alexus [3.1K]

Answer:

y=4/5+3 and 3/5

Step-by-step explanation:

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2 years ago

Peggy earned $20 for each lawn she mowed last summer. This summer, she raised her price to $23 per lawn. What is the percent inc

anyanavicka [17]

Answer:

15 percent

Step-by-step explanation: You will substract the new value from the old value.

23-20

Now you will get the difference between them and divide that by the original amount.

3/20

This will equal 0.15 in which you now multiply by 100

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2 years ago

Use the distance formula distance, to find the distance to the nearest tenth , between each pair of points A(6,2) and D(-3,-2)

marusya05 [52]

Answer:

The distance between A and D to the nearest tenth is;

$9.8\text{ units}$

Explanation:

Given the two points;

$A(6,2)\text{ and D(-3,-2)}$

Applying the distance between two points formula;

$d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}$

substituting the given coordinates we have;

$\begin{gathered} AD=\sqrt[]{(-3-6)^2+(-2-2)^2} \\ AD=\sqrt[]{(-9)^2+(-4)^2} \\ AD=\sqrt[]{81+16} \\ AD=\sqrt[]{97} \end{gathered}$

Simplifying;

$\begin{gathered} AD=9.8488578 \\ AD\approx9.8 \end{gathered}$

Therefore, the distance between A and D to the nearest tenth is;

$9.8\text{ units}$

4 0

1 year ago

This is a 3 part 1 question each part on how to locate the plane and a small explanation report working out the recovery details

sesenic [268]

Third leg.

The crew flies at a speed of 560 mi/h in direction N-20°-E.

The wind has a speed of 35 mi/h and a direction S-10°-E.

We then can draw this as:

We have to add the two vectors to find the actual speed and direction.

We will start by adding the x-coordinate (W-E axis):

$\begin{gathered} x=560\cdot\sin (20\degree)+35\cdot\sin (10\degree) \\ x\approx560\cdot0.342+35\cdot0.174 \\ x\approx191.53+6.08 \\ x\approx197.61 \end{gathered}$

and the y-coordinate (S-N axis) is:

$\begin{gathered} y=560\cdot\cos (20\degree)-35\cdot\cos (10\degree) \\ y\approx560\cdot0.940-35\cdot0.985 \\ y\approx526.23-34.47 \\ y\approx491.76 \end{gathered}$

Then, the actual speed vector is v3=(197.61, 491.76).

The starting location for the third leg is R2=(216.66, 167.67) [taken from the previous answer].

Then, we have to calculate the displacement in 20 minutes using the actual speed vector.

We can calculate the movement in each of the axis. For the x-axis:

$\begin{gathered} R_{3x}=R_{2x}+v_{3x}\cdot t \\ R_{3x}=216.66+197.61\cdot\frac{1}{3} \\ R_{3x}=216.66+65.87 \\ R_{3x}=282.53 \end{gathered}$

NOTE: 20 minutes represents 1/3 of an hour.

We can do the same with the y-coordinate:

$\begin{gathered} R_{3y}=R_{2y}+v_{3y}\cdot t \\ R_{3y}=167.67+491.76\cdot\frac{1}{3} \\ R_{3y}=167.67+163.92 \\ R_{3y}=331.59 \end{gathered}$

The final position is R3 = (282.53, 331.59).

To find the distance from the origin and direction, we transform the cartesian coordinates of R3 into polar coordinates:

The distance can be calculated as if it was a right triangle:

$\begin{gathered} d^2=x^2+y^2_{} \\ d^2=282.53^2+331.59^2 \\ d^2=79823.20+109951.93 \\ d^2=189775.13 \\ d=\sqrt[]{189775.13} \\ d\approx435.63 \end{gathered}$

The angle, from E to N, can be calculated as:

$\begin{gathered} \tan (\alpha)=\frac{y}{x} \\ \tan (\alpha)=\frac{331.59}{282.53} \\ \tan (\alpha)\approx1.1736 \\ \alpha=\arctan (1.1736) \\ \alpha=49.56\degree \end{gathered}$

If we want to express it from N to E, we substract the angle from 90°:

$\beta=90\degree-\alpha=90-49.56=40.44\degree$

Answer: the final location can be represented with the vector (282.53, 331.59).

1) The distance from the origin is 435.63 miles and

2) the direction is N-40°-E.

7 0

1 year ago