All categories

3 years ago

Please answer please please answer

Mathematics

1 answer:

cestrela7 [59]3 years ago

3 0

Answer:

Step-by-step explanation:

i could be wrong sorry in advance if it is i am not 100 percent sure

You might be interested in

1. (a) Let A = 1, 3, 5, 7 and B = X : X is the first three multiple of 7 . Find (A−B).

ICE Princess25 [194]

Answer:

a) A-B = { 1 , 3, 5 }

b) The remainder is 50

c) f + g)(x) = x⁴+ 5x³

d) $\frac{dy}{dx} = \frac{2x- y^3(3x^2)) }{ 1- x^3 3y^2}$

Step-by-step explanation:

Step(I):-

1) a)

Let A = { 1, 3, 5, 7 } and B = { 7 , 14 , 21 }

A - B = {1,3,5,7} - { 7,14,21}

A - B = { 1 , 3, 5}

Given f (x)=3x³−11x−48

x-1 ) 3 x³ - 11x -48 (- 3x²-3x -2

-3x³ +3x²

3x²-11x-48

-3x²+9x

2x -48

-2x +2

The remainder is 50

c) Given f(x) = x³ and g(x) = x + 5

(f. g)(x) = f(x) .g(x) = x³ . (x+5) = x³(x) + 5 x³ = x⁴+ 5x³

Given function y = 3 + x²+x³y³ ...(I)

By using derivative formulas

y = xⁿ

$\frac{dy}{dx} = n x^{n-1}$

Apply d/dx ( UV) = U V¹ + V U¹

Differentiating equation (I) with respective to 'x' , we get

$\frac{dy}{dx} = 0 + 2x + ( x^3 3y^2(\frac{dy}{dx} )+ y^3(3x^2))$

$\frac{dy}{dx} - ( x^3 3y^2(\frac{dy}{dx} )= 2x- y^3(3x^2))$

$\frac{dy}{dx} = \frac{2x- y^3(3x^2)) }{ 1- x^3 3y^2}$

7 0

2 years ago

Find the missing angle

sineoko [7]

129 degrees is the correct answer for the missing angle number 1 :)

3 0

2 years ago

a town Pharmacy in hospital is doing it as a part of format of the town has a scale 18840.35 miles what is the distance between

allsm [11]

Answer:34344e4365655.6

Step-by-step explanation:

3 0

3 years ago

Due later pls help

Andre45 [30]

Answer:

1. 4z=-8

z= -8/4= -2

2. 2/3t=8/1

2= 24t

t= 0.08

3. 15-1=-8d

14=-8d

d= -1.75

4. 5r-2r=18

3r= 18

r= 6

5. 24= 8y

y=3

6. 3t-3= -t+5

3t+t= 5+3

4t=8

t=2

7. 2x-x= 3+6

x=9

5 0

3 years ago

The triangle described is of the form SSA. Determine if there is no triangle possible, one triangle possible, or two triangles p

kvv77 [185]

Answer: The way to work this problem is to sketch the SSA-described triangle.

Horizontally, mark a ray stemming from point C, going to the right.

From point C, at an angle of elevation of 20 degrees, draw a segment a = 6.3, ending at point B.

Now, from point B, draw a segment perpendicular to the horizontal ray and label it h, for altitude.

Calculate h = 6.3 sin 20o = 2.1547.

Now imagine that, dangling from point B, you have a segment of length c = 9.3; in fact, put a compass with steel point at B and radius 9.3, and draw a big arc. you'll see that it can intersect the horizontal ray only once; in its swing toward point C, due to c > a, there will be no second intersection of the ray (the arc 'overshoots' the ray).

Thus you see that only one solution exists, and the Law of Sines can be used to solve the triangle:

sin 20 / 9.3 = sin A / 6.3 and this leads to A = 13.4 degrees...and B = 180 - 20 - 13.4 = 146o.

I'll let you solve for b (side AC).

6 0

3 years ago