All categories

2 years ago

I was never taught how to do this... some one help me.

Mathematics

2 answers:

Tatiana [17]2 years ago

8 0

So cos means adjacent over hypotenuse which would be 15/17 and you can’t simplify it down farther so 15/17 is your final answer.

lys-0071 [83]2 years ago

6 0

$\cos X = \frac{\text{adjacent}}{\text{hypotenuse}} = \boxed{\frac{15}{17}}$

You might be interested in

Please help with this!

shepuryov [24]

9514 1404 393

Answer:

-11

Step-by-step explanation:

DeYanna's cards are ...

black: 2, 4, for a total value of 2+4 = 6

red: 7, 10, for a total value of -(7+10) = -17

Then the total value of DeYanna's cards is ...

6-17 = -11

7 0

3 years ago

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago

(-2h3n2)(-5h2n2)<br> How

IRINA_888 [86]

Answer: 9h x 8n^2

we rearrange this underneath the question (-5h-2h) (2nx3n)(2+2) or if 2^2 see explanation. It must therefore be 7h6n4

Step-by-step explanation:

if its ^2 )(power2) then the second reasonable answer is (7h-2h)(2n^2 x 2n^2)=9h x 4n^2 x4n^2 =9h 8n^2 or 17hn^2 but i leave as 9h 8n^2

You can expand only if multiplying 9hx4 for example but not combine the letters. The questions that ask you to mix are much different to this. but if you want to add them 9h+8n^2 =17hn^2

3 0

3 years ago

Factor:14x3 – 21x

Vlada [557]

7x (2x^2 - 3) is your answer my dude/lady

7 0

3 years ago

What is the circumference of a circle with a diameter of 190mm?

MrMuchimi

Answer:

it's 502.6548246

Step-by-step explanation:

hhhhhhhg

7 0

3 years ago