SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!

What height on the wall will a 15 foot ladder reach if it is placed 3.5 feet from the base of the wall...???

umka2103 [35]

Answer:

14.6 feet

Step-by-step explanation:

Let the height of the ladder be the hypotenuse, the distance from the wall be the base and the height till the ladder reaches be the perpendicular.

Using Pythogoras theorem,

H^2 = p^2+b^2

p^2 = H^2-b^2

p = under root(H^2-b^2)

p = under root(225 - 12.25)

p = 14.6 feet

7 0

3 years ago

Read 2 more answers

Given the system of equations, what is the solution? 3x-2y+10=0 5y=4x+8

masha68 [24]

X = -34/7
Y = -16/7
I need to add words to get to 20 characters

8 0

2 years ago

The middle 5 in 0.555 is 1/10 the value of the 5 to its _____

Molodets [167]

Right. The first 5 is in the tenths place, the second five is in the hundredths, and the third one is in the thousandths. I'd like to think that if you take away (divide 1000 by ten) a tenth from the thousandths, that would leave you with the 5 in the middle being one tenth the value of the five to its right. (I'm not sure if this is correct or not, i'm just making an assumption). I would check with someone to make sure

7 0

3 years ago

Read 2 more answers

A small plane and a large plane are 6.8km from each other, at the same altitude (height). From an observation tower, the two air

Lemur [1.5K]

Answer:

7.9km

Step-by-step explanation:

(a)See attached for the diagram representing this situation.

(b)

In Triangle ABC

$\text{Using Law of Sines}\\\dfrac{\sin A}{a}=\dfrac{\sin C}{c} \\\dfrac{\sin A}{5.2}=\dfrac{\sin 58^\circ}{6.8} \\\sin A=5.2 \times \dfrac{\sin 58^\circ}{6.8}\\A=\arcsin (5.2 \times \dfrac{\sin 58^\circ}{6.8})\\A=40.43^\circ$

Next, we determine the value of Angle B.

$\angle A+\angle B+\angle C=180^\circ\\40.43+58+\angle B=180^\circ\\\angle B=180^\circ-(40.43+58)\\\angle B=81.57^\circ$

Finally, we find b.

$\text{Using Law of SInes}\\\dfrac{b}{\sin B}=\dfrac{c}{\sin C} \\\dfrac{b}{\sin 81.57^\circ}=\dfrac{6.8}{\sin 58^\circ} \\b=\dfrac{6.8}{\sin 58^\circ} \times \sin 81.57^\circ\\b=7.9km $ (to the nearest tenth of a kilometer)$

The distance between the small plane and the observation tower is 7.9km.

8 0

3 years ago

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th

iris [78.8K]

Answer:

t = 15.42 s

Step-by-step explanation:

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation.

$y=-16x^2+242x+73$

We ned to find the time when the rocket will the ground. For this, y = 0

$-16x^2+242x+73=0$

It is a quadratic equation whose solution is given by :

$x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

Here, a = -16, b = 242 and c = 73

$x=\dfrac{-b+ \sqrt{b^2-4ac} }{2a}, x=\dfrac{-b-\sqrt{b^2-4ac} }{2a}\\\\x=\dfrac{-242+ \sqrt{(242)^2-4(-16)(73)} }{2(-16)}, x=\dfrac{-242- \sqrt{(242)^2-4(-16)(73)} }{2(-16)}\\\\=-0.295\ s, 15.42\ s$

Neglecting negative value.

So, it will hit the ground in 15.42 s.

5 0

3 years ago