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3 years ago

11

I'm begging you please please help please please ASAP please please help please please ASAP please please help please please ASA

P please please help please please ASAP

1 answer:

Oksanka [162]3 years ago

6 0

Answer:

15

Step-by-step explanation:

the height of the person is y = 6 and the distance from the ladder is also 6 = x. To find the angle formed by the ladder we use $arctan(\frac{y}{x})$ which in this case is

$arctan\frac{6}{6} = arctan(1)\\= 45 degrees$

now we know that the person is 9 feet away from wall and 6 feet away from the ladder so the total x distance is 9+6 = 15 = x

to find y we use tan(45) = $\frac{y}{x}$

multiply both sides by x = 15 and simplify to get y = 15

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12.5% of 360 is 45
360-45 is 315
Answer is £315

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3 years ago

At 3:25 p.m., a passenger train and a freight train both left Kalamazoo,

Svetlanka [38]

Answer:

4:10 PM

Step-by-step explanation:

8 0

3 years ago

What is the answer to this problem?

Slav-nsk [51]

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3 years ago

Read 2 more answers

The set of points (-3, 4), (-1, 1), (-3,-2), and (-5,1) identifies the vertices of a quadrilateral. Which is the most specific d

KIM [24]

Answer:

Option (4). Rhombus

Step-by-step explanation:

From the figure attached,

Distance AB = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

= $\sqrt{(1-4)^2+(-5+3)^2}$

= $\sqrt{(-3)^2+(-2)^2}$

= $\sqrt{13}$

Distance BC = $\sqrt{(4-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance CD = $\sqrt{(-2-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance AD = $\sqrt{(1+2)^2+(-5+3)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Slope of AB ( $m_{1}$ ) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{4-1}{-3+5}$

= $\frac{3}{2}$

Slope of BC ( $m_{2}$ ) = $\frac{4-1}{-3+1}$

= $-\frac{3}{2}$

If AB and BC are perpendicular then,

$m_{1}\times m_{2}=-1$

But it's not true.

[ $m_{1}\times m_{2}=(\frac{3}{2})(-\frac{3}{2})$ = - $\frac{9}{4}$ ]

It shows that the consecutive sides of the quadrilateral are not perpendicular.

Therefore, ABCD is neither square nor a rectangle.

Slope of diagonal BD = $\frac{4+2}{-3+3}$

= Not defined (parallel to y-axis)

Slope of diagonal AC = $\frac{1-1}{-1+5}$

= 0 [parallel to x-axis]

Therefore, both the diagonals AC and BD will be perpendicular.

And the quadrilateral formed by the given points will be a rhombus.

5 0

3 years ago

Plssss help me it’s really important!!!

slavikrds [6]

Answer:

B

Step-by-step explanation:

Since ∠ E = ∠ F then the triangle is isosceles and DF = DE, that is

6x = x + 2y ( subtract x from both sides )

5x = 2y ( subtract 2y from both sides )

5x - 2y = 0 → (1)

and perimeter = 29, that is

6x x + 2y + 5x - y = 29 , simplifying

12x + y = 29 → (2)

Multiply (2) by 2

24x + 2y = 58 → (3)

Add (1) and (3) term by term to eliminate y, that is

29x = 58 ( divide both sides by 29 )

x = 2

Substitute x = 2 into (2) and evaluate for y

12(2) + y = 29

24 + y = 29 ( subtract 24 from both sides )

y = 5

Thus

FE = 5x - y = 5(2) - 5 = 10 - 5 = 5 → B

3 0

3 years ago