Solve the equation for (x) and simplify. (7X/6) = (5//2),x

Is it possible for a system of linear equations to have no solutions ?

sveta [45]

Yes it is possible,they never intersect, so there is no point that lies on both lines, and no solution to the system

8 0

3 years ago

Read 2 more answers

F(x) = 5x2 – 20x + 26 In vertex form

mihalych1998 [28]

Answer:

$f(x)=5(x-2)^{2}+6$

Step-by-step explanation:

we have

$f(x)=5x^{2} -20x+26$

This is the equation of a vertical parabola open upward

The vertex is a minimum

The equation of a vertical parabola in vertex form is equal to

$f(x)=a(x-h)^2+k$

where

(h,k) is the vertex

Convert the given function to vertex form

Factor the leading coefficient

$f(x)=5(x^{2} -4x)+26$

Complete the square

$f(x)=5(x^{2} -4x+4)+26-20$

$f(x)=5(x^{2} -4x+4)+6$

Rewrite as perfect squares

$f(x)=5(x-2)^{2}+6$ ------> equation in vertex form

The vertex is the point (2,6)

6 0

3 years ago

8. Find the area and the perimeter of

shtirl [24]

Answer:

the answer is DBC=90°

because the midpoint of the perimeter of quadrilateral is B

4 0

2 years ago

Find the vector and parametric equations for the line through the point P(0, 0, 5) and orthogonal to the plane −1x+3y−3z=1. Vect

yulyashka [42]

Answer:

Note that orthogonal to the plane means perpendicular to the plane.

Step-by-step explanation:

-1x+3y-3z=1 can also be written as -1x+3y-3z=0

The direction vector of the plane -1x+3y-3z-1=0 is (-1,3,-3).

Let us find a point on this line for which the vector from this point to (0,0,5) is perpendicular to the given line. The point is x-0,y-0 and z-0 respectively

Therefore, the vector equation is given as:

-1(x-0) + 3(y-0) + -3(z-5) = 0

-x + 3y + (-3z+15) = 0

-x + 3y -3z + 15 = 0

Multiply through by - to get a positive x coordinate to give

x - 3y + 3z - 15 = 0

4 0

2 years ago

Can anyone help me with these i need them befire tomorrow

mash [69]

When you have a scale factor as percentage, is better to convert that percentage into a fraction by divide it by 100%. That way, it will be much easier to scale your figures.

1. The scale factor we need to apply to our rectangle is 60%. Firts, lets convert our scale factor into a fraction: $\frac{60}{100} = \frac{3}{5}$ . Next, we just need to multiply each side of our rectangle by our fraction:
length= $16( \frac{3}{5})=9.6$
width= $8( \frac{3}{5} )=4.8$
So, our new rectangle will have a length of 9.6 units and a width of 4.8 units.

2. This time the scale factor is 120%. Just as before, lets convert it into a fraction: $\frac{120}{100} = \frac{6}{5}$ , and just as before lets multiply each side of our triangle by our fraction:
base= $10( \frac{6}{5} )=12$
side= $8( \frac{6}{5} )=9.6$
So, our new isosceles triangle will have a base of 12 units and tow sides of length 9.6 units each.

3. This time the scale factor is 150%. Just as before, lets convert it into a fraction: $\frac{150}{100} = \frac{3}{2}$ , and just as before lets multiply each side of our polygon by our fraction:
$12( \frac{3}{2} )=18$
$15( \frac{3}{2} )=22.5$
$13( \frac{3}{2} )=19.5$
So, our new polygon will have a side of 12 length units, tow sides of 15 length units, and two side of 15 length units.

4. This time our factor is 75%. Just as before, lets convert it into a fraction: $\frac{75}{100} = \frac{3}{4}$ , and just as before lets multiply each side of our parallelogram by our fraction:
side= $25( \frac{3}{4} )=18.75$
So, our new parallelogram will have tow sides of length 18.75 units.

7 0

3 years ago