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3 years ago

-56 + _ =-84 please help

Mathematics

1 answer:

GaryK [48]3 years ago

4 0

Answer:

-28 is the answer.

Step-by-step explanation:

84-56=28

56+28=84

-56+-28=-84

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3. What is the area of the triangle below? Round your

Jobisdone [24]

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3 years ago

What is the slope of the line shown below (- 1, 8) (2, - 4)

suter [353]

Answer:

The slope of the line can be determined from the gradient equation, rise (y axis)/ run (x-axis)

Hence let (-1, 8) be A and (2, -4) be B. The slope from A to B is (-4-8)/(2-(-1))= (-12)/3= -4

Hence the gradient/ slope of the line is -4.

Step-by-step explanation:

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3 years ago

the sum of four times the first number and three times the second number is 15. the difference of three times the first number a

Nana76 [90]

Hey there!

Let's create a systems of equations using x and y.

4x+3y=15

3x-2y=7

We need to solve this using elimination. First, we need to multiply the first equation by 2/3 so that we can cancel out the y's when combining our equations.

2 2/3x+2y=10

3x-2y=7

Now we combine the equations...

5 2/3x=17

x=3

Now we can plug our x into the first equation to find y.

12+3y=15

y=1

Our numbers are 3 and 1.

I hope this helps!

3 0

2 years 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

7 months ago

Need help for test asap!

klio [65]

Answer:

do the test then

Step-by-step explanation:

do the test then

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2 years ago