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2 years ago

What is the value of x ??

Mathematics

1 answer:

sineoko [7]2 years ago

4 0

Answer:

Step-by-step explanation:

The cosine of an acute angle in a right triangle is the ratio of the length of the adjacent side to the length of the hypotenuse.

$\tt \cos A=\cfrac{x}{AC} \ \ \Rightarrow \ \ x=AC\cdot\cos A=24\sqrt{2}\cdot\cfrac{\sqrt{2}}{2}=24 \ units$

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2 years ago

The universal set is U = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}. If A = {Monday, Tuesday,Wednesday, Th

Varvara68 [4.7K]

Answer:

{Monday, Tuesday, Wednesday, Thursday}

Step-by-step explanation:

First we need to get the complement of B i.e B'

B' are the set in the Universal set but that are not in B

B' = {Monday, Tuesday, Wednesday, Thursday}

A= {Monday, Tuesday, Wednesday, Thursday, Friday}

A intersect B ' are the days of the week that are common to both sets;

A intersect B ' = {Monday, Tuesday, Wednesday, Thursday}

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1 year ago

In an agricultural class, there are 17 freshmen students and 19 sophomore students. What ratio compares the number of freshmen t

zheka24 [161]

Answer:

B. F:S = 17 to 19; S:F = 19 to 17

Step-by-step explanation:

Put the corresponding numbers in the corresponding parts of the ratio you want to write:

freshmen to sophomores (F:S)

17 to 19

___

sophomores to freshmen (S:F)

19 to 17

These pairs of numbers match choice B.

5 0

2 years ago

Does anyone know the answer???

Fiesta28 [93]

Answer:

-2

Step-by-step explanation:

41 - 8m - 57 = 0

41- 8m = 57

-8m = 57-41

-8m = 16

m = -16/8

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8 0

2 years ago

Read 2 more answers

Use Lagrange multipliers to find the maximum and minimum values of

marusya05 [52]

The Lagrangian is

$L(x,y,\lambda)=x+8y+\lambda(x^2+y^2-4)$

It has critical points where the first order derivatives vanish:

$L_x=1+2\lambda x=0\implies\lambda=-\dfrac1{2x}$

$L_y=8+2\lambda y=0\implies\lambda=-\dfrac4y$

$L_\lambda=x^2+y^2-4=0$

From the first two equations we get

$-\dfrac1{2x}=-\dfrac4y\implies y=8x$

Then

$x^2+y^2=65x^2=4\implies x=\pm\dfrac2{\sqrt{65}}\implies y=\pm\dfrac{16}{\sqrt{65}}$

At these critical points, we have

$f\left(\dfrac2{\sqrt{65}},\dfrac{16}{\sqrt{65}}\right)=2\sqrt{65}\approx16.125$ (maximum)

$f\left(-\dfrac2{\sqrt{65}},-\dfrac{16}{\sqrt{65}}\right)=-2\sqrt{65}\approx-16.125$ (minimum)

5 0

2 years ago