Ask question

Ask question

All categories

3 years ago

8

Pls help and explain

1 answer:

Alexeev081 [22]3 years ago

4 0

Answer:

The height of the triangle is 28 inches.

Step-by-step explanation:

308 = (h * l) / 2

308 = (h * 22) / 2

616 = 22h

h = 28

You might be interested in

Which classification describes the system of linear equations?

Aleksandr [31]

Answer:

A system of two linear equations can have one solution, an infinite number of solutions, or no solution. Systems of equations can be classified by the number of solutions. If a system has at least one solution, it is said to be consistent . If a consistent system has exactly one solution, it is independent .

Step-by-step explanation:

3 0

1 year ago

$2,000 loan at 8% for 5 years.

Harman [31]

<u>Answer:</u>

$2,800 = Principle + Interest

<u>Step-by-step explanation:</u>

Using PRT/100, let's find the interest.

=> 2,000 x 8 x 5/100
=> 20 x 8 x 5
=> $800

<u>Conclusion: </u>

Therefore, the interest plus principle is 2,000 + 800, which is $2,800.

Hoped this helped.

$BrainiacUser1357$

8 0

2 years ago

Read 2 more answers

Determine the slope-intercept form of the equation of the line parallel to y = x + 11 that passes through the point (–6, 2).

KATRIN_1 [288]

Answer:

y=4x+6

Step-by-step explanation:

4 0

3 years ago

What is a function? and is this a function?

hichkok12 [17]

Answer: it is not a function if the x repeats

Step-by-step explanation: My teacher always says THE X DOES NOT REPEAT that is the easiest way to remember it

8 0

3 years ago

Read 2 more answers

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

3 years ago