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sveta [45]
3 years ago
6

What is the cosine ratio of ∠Z?

Mathematics
2 answers:
rjkz [21]3 years ago
4 0

Answer:

Cos of an angle by definition of its ratio is side adjacent/hypotenuse.  The side adjacent to angle Z cannot be the hypotenuse, so it has to be 10. he hypotenuse is 26.  Therefore, cosZ= 10/26 = 5/13

Step-by-step explanation:

MRK ME BRAINLIEST PLZZZZZZZZZZZZz

Ierofanga [76]3 years ago
3 0

Answer:

10 / 26

Step-by-step explanation:

cos <Z = adjacent / hypotenuse

= YZ / XZ

= 10 / 26

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A boat travels upstream for 2 hours. the return trip only takes 1.7 miles because the boat travels 2.5 miles faster downstream.
Ilia_Sergeevich [38]

Answer:

the boat travel one way = 2 x 14.16 = 28.32 miles

Step-by-step explanation:

the return trip only takes 1.7 miles (?) ... is it 1.7 hours (?)

the boat travels 2.5 miles (per hour ?) faster downstream

I calculate it on the assumption of my guess stated above ...

X: represent the boat speed / hour

distance one way = 2X

and Distance also = 1.7 * (x + 2.5 ) on return

2x = 1.7 x + 1.7 * 2.5 = 1.7x + 4.25

0.3 x = 4.25

x = 14.16

the boat travel one way = 2 x 14.16 = 28.32 miles

3 0
3 years ago
A random sample of 95 vehicles is taken from a large parking lot at an office park. Below is the type of each vehicle, and wheth
kati45 [8]

Answer:

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Step-by-step explanation:

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4 years ago
Use the procedures developed to find the general solution of the differential equation. (Let x be the independent variable.)
makkiz [27]

Solution :

Given :

2y''' + 15y'' + 24y' + 11y= 0

Let x = independent variable

(a_0D^n + a_1D^{n-1}+a_2D^{n-2} + ....+ a_n) y) =  Q(x)  is a differential equation.

If Q(x) \neq 0

It is non homogeneous then,

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If Q(x) = 0

It is called the homogeneous then the general solution =  complementary solution.

2y''' + 15y'' + 24y' + 11y= 0

$(2D^3+15D^2+24D+11)y=0$

Auxiliary equation,

$2m^3+15m^2+24m +11 = 0$

-1  | 2    15    24     11

   <u> | 0   -2    - 13    -11  </u>

      2    13    11       0

∴ 2m^2+13m+11=0

The roots are

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

$=\frac{-13\pm \sqrt{13^2-4(11)(2)}}{2(2)}$

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So, m_1, m_2, m_3 = -1, -1, -5.5

Then the general solution is :

$= (c_1+c_2 x)e^{-x} + c_3 \ e^{-5.5x}$

 

8 0
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oksian1 [2.3K]

Answer:

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Step-by-step explanation:

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