All categories

2 years ago

What is a - 2 = 3 - a/2 + 4 How i do i get the answer

Mathematics

1 answer:

slamgirl [31]2 years ago

8 0

Answer:

a=6

Step-by-step explanation:

$a-2=3-\frac{a}{2}+4 \\ \\ a-2=-\frac{a}{2}+7 \\ \\ 2a-4=-a+14 \\ \\ 3a=18 \\ \\ a=6$

You might be interested in

Which point on the number line represents the product of 4 and –2?

Goshia [24]

Answer:

It's A

Step-by-step explanation:

the one above me is incorrect it's "A"

5 0

3 years ago

Read 2 more answers

Find a point on the ellipsoid x2+y2+4z2=36x2+y2+4z2=36 where the tangent plane is perpendicular to the line with parametric equa

Anvisha [2.4K]

Answer:

A point on the ellipsoid is (-4,2,2) or (4,-2,-2)

Step-by-step explanation:

Given equation of ellipsoid f(x,y,z) : $x^2+y^2+4z^2=36$

Parametric equations:

x=-4t-1

y=2t+1

z=8t+3

Finding the gradient of function

$\nabla f(x,y,z)=\\\nabla f(x,y,z)=$

So, The directions vectors=(-4,2,8)

Now the line is perpendicular to plane when direction vector is parallel to the normal vector of line

$\nablaf(x,y,z)=(2x,2y,8z)=\lambda(-4,2,8)$

So, $2x=-4\lambda$

$\Rightarrow x=-2\lambda$

$2y=2\lambda\\\Rightarrow y=\lambda\\8z=8\lambda\\\Rightarrow z=\lambda$

Substitute the value of x , y and z in the ellipsoid equation

$(2\lambda)^2+(\lambda)^2+4(\lambda)^2=36\\9(\lambda)^2=36\\\lambda^2=4\\\lambda=\pm 2$

With $\lambda = 2$

x=-2(2)=-4

y=2

z=2

With $\lambda =- 2$

x=-2(-2)=4

y=-2

z=-2

Hence a point on the ellipsoid is (-4,2,2) or (4,-2,-2)

7 0

4 years ago

URGENT! Can anyone explain how to solve this??

Alex

Check the picture below

the triangles are similar, the angles are congruents then

so... just add them up, and divide by 2, or set them as rational

7 0

3 years ago

If f(x-3)=x^2-4x+5, what's f(1-x)

olganol [36]

Answer:

$\large\boxed{f(1-x)=x^2-4x+5}$

Step-by-step explanation:

$f(x-3)=x^2-4x+5\\\\f((x-3)+3)\to\text{put x + 3 to the equation of the function:}\\\\f(x-3+3)=(x+3)^2-4(x+3)+5\\\text{use the distributive property and}\ (a+b)^2=a^2+2ab+b^2\\f(x)=x^2+2(x)(3)+3^2+(-4)(x)+(-4)(3)+5\\f(x)=x^2+6x+9-4x-12+5\qquad\text{combine like terms}\\f(x)=x^2+(6x-4x)+(9-12+5)\\f(x)=x^2+2x+2\\\\f(1-x)\to\text{put 1 - x to the equation of the funtion f(x):}\\\\f(1-x)=(1-x)^2+2(1-x)+2\\\text{use the distributive property and}\ (a-b)^2=a^2-2ab+b^2$

$f(1-x)=1^2-2(1)(x)+x^2+(2)(1)+(2)(-x)+2\\f(1-x)=1-2x+x^2+2-2x+2\qquad\text{combine like terms}\\f(1-x)=x^2+(-2x-2x)+(1+2+2)\\f(1-x)=x^2-4x+5$

4 0

4 years ago

Math help please TY :3<br><br> HWAJDSK 10 POINTS REWARDING !!

valkas [14]

I will do Point A carefully, The others I will indicate. Start with the Given Point A. Then do the translations

A(-1,2) Original Point

Reflection: about x axis:x stays the same; y becomes -y:Result(-1,-2)

T<-3,4>: x goes three left, y goes 4 up (-1 - 3, -2 + 4): Result(-4,2)

R90 CCW: Point (x,y) becomes (-y , x ) So (-4,2) becomes(-2, - 4): Result (-2, - 4)

B(4,2) Original Point

Reflection: (4, - 2)
T< (-3,4): (4-3,-2 + 4): (1 , 2)
R90 CCW: (-y,x) = (-2 , 1)

C(4, -5) Original Point

Reflection (4,5)
T<-3,4): (4 - 3, 5 + 4): (1,9)
R90, CCW (-9 , 1)

D(-1 , -5) Original Point

Reflection (-1,5)
T(<-3,4): (-1 - 3, 5 + 4): (-4,9)
R90, CCW ( - 9, - 4)

Note: CCW means Counter Clockwise

The graph on the left is the same one you have been given.

The graph on the right is the same figure after all the transformations

7 0

3 years ago