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3 years ago

Please guys help me out . Thanks

Mathematics

1 answer:

Naya [18.7K]3 years ago

6 0

Answer:

6.7

Step-by-step explanation:

cos(48) = adjacent/hypotenuse

cos(48) = x/10

x = 10cos(48)

x = 6.691306064

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Give the degree of the polynomial <br> -3w+x^6y^5-2+5y^4w^3x^2

Jlenok [28]

Answer:

Here the degree of the polynomial is 11.

Step-by-step explanation:

To find the degree of the multivariate polynomials, we need to add up the powers of all the variables. So the total degree is given by the sum of all the powers of the highest powers terms.

Now in the given polynomial $-3w+x^6y^5-2+5y^4w^3x^2$

the term with the highest total powers is $x^6y^5$ and thus the total power is 6+5=11.

And hence the degree of this polynomial is 11.

5 0

3 years ago

Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel

nydimaria [60]

Answer with Step-by-step explanation:

We are given that

F=<0,-8>=0i-8j=-8j

$\theta=\frac{\pi}{3}$

The component of force is divided into two direction

1.Along the plane

2.Perpendicular to the plane

1.The vector parallel to the plane will be= $r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j$

By using $cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}$

Force along the plane will be= $\mid F_x\mid=F\cdot r$

Force along the plane will be = $\mid F_x\mid=F\cdot (\frac{1}{2}i-\frac{\sqrt 3}{2}j)=-8j\cdot(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=8\times \frac{\sqrt 3}{2}=4\sqrt 3$ N

By using $i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0$

Therefore, force along the plane= $\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)$

2.The vector perpendicular to the plane= $r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j$

The force perpendicular to the plane= $\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

The force perpendicular to the plane= $4$ N

Therefore, $F_y=4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $2\sqrt 3i-6j-2\sqrt3 i-2j=-8j$

Hence,sum of two component of forces=Total force.

6 0

3 years ago

Please help ASAP! 40 PTS

tester [92]

Answer:

see below

Step-by-step explanation:

1. < 4 is the exterior angle of a triangle given

2. <1 + <2 + <3 =180 basic property of triangles

3. <3 + <4 are a linear pair <3 and <4 are adjacent and form a straight line

4. <3 and <4 are supplementary angles supplementary angles are linear pairs

5. <3 + <4 = 180 The definition of supplementary angles are that they add to 180 degrees

6. <1 + <2 + <3 = <3+ <4 Substitute <3 + <4 for 180 in equation from step 2

7 <1 + <2 = <4 Subtract <3 from each side using the subtraction property of equality

4 0

3 years ago

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Solve for x and y<br> 6(ax + by) = 3a + 2b ; 6(bx-ay) = 3b - 2a

slava [35]

6×ax+6×by =6ax+6by =3a+2b,6×bx-6×ay=3b-2a=6bx-3a+2b

4 0

2 years ago

How is a 15% discount similar to a 15% decrease? Explain

VMariaS [17]

It's the same because you're trying to find net price (end price)

You're going to take in both scenarios 100%-15%= 85%

You're always going to be paying only 85% of the price.

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3 years ago

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