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3 years ago

HELP ASAP I DONT HAVE TIME IT ALSO DETECTS IF ITS RIGHT OR WRONG

Mathematics

1 answer:

storchak [24]3 years ago

5 0

Answer:

9/2*1/6=0.75

Step-by-step explanation:

hope this helped!!!!

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PLEASE HELP!!! DUE IN 10 MIN!!!

9966 [12]

Answer:

Option B

Step-by-step explanation:

Surface area of a triangular prism = Area of the triangular sides + Area of the rectangular sides

Area of the triangular sides = 2(Area of the triangular base)

= $2[\frac{1}{2}(\text{Base})(\text{Height})]$

= 10 × 12

= 120 ft²

Area of the rectangular sides = (Perimeter of the triangular side)(Height of the prism)

= (10 + 13 + 13)(15)

= 36 × 15

= 540 ft²

Surface area of the prism = 120 + 540

= 660 square feet

Option B will be the answer.

5 0

3 years ago

Find a quadratic polynomial with integer coefficients which has x = 2/9, ± SQRT23/9 as its real zeros.

melomori [17]

Answer:

The quadratic polynomial with integer coefficients is $y = 81\cdot x^{2}-36\cdot x -19$ .

Step-by-step explanation:

Statement is incorrectly written. Correct form is described below:

Find a quadratic polynomial with integer coefficients which has the following real zeros: $x = \frac{2}{9}\pm \frac{\sqrt{23}}{9}$ .

Let be $r_{1} = \frac{2}{9}+\frac{\sqrt{23}}{9}$ and $r_{2} = \frac{2}{9}-\frac{\sqrt{23}}{9}$ roots of the quadratic function. By Algebra we know that:

$y = (x-r_{1})\cdot (x-r_{2}) = x^{2}-(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2}$ (1)

Then, the quadratic polynomial is:

$y = x^{2}-\frac{4}{9}\cdot x -\frac{19}{81}$

$y = 81\cdot x^{2}-36\cdot x -19$

The quadratic polynomial with integer coefficients is $y = 81\cdot x^{2}-36\cdot x -19$ .

5 0

3 years ago

Yall I need help QUICK

Tomtit [17]

Answer:

y=-3x-12

Step-by-step explanation:

6 0

3 years ago

What is the vertex form, f(x) = a(x - h)2-k, for a parabola that passes through the point (2,8) and has (1,0) as its

Novay_Z [31]

Answer:

Step-by-step explanation:

hello :

f(x) = a(x - h)²-k ....vertex form when the vertex is : (h,k)

h=1 and k=0 so : f(x) = a(x-1)²

a parabola that passes through the point (2,8) : f(2)=8

a(2-1)² =8

a= 8

f(x) = 8(x-1)²= 8(x²-2x+1)

f(x) 8x²-16x+8 .....standard form

3 0

3 years ago

Find parametric equations through point P=(2,2,7) in the direction of the vector v = (44, 14, -20)?

ycow [4]

9514 1404 393

Answer:

(x, y, z) = (2+44t, 2+14t, 7-20t)

Step-by-step explanation:

One way to write parametric equations for line L is ...

L = P + t·v

where P is the given point and v is the given direction vector. Using that form, we have ...

(x, y, z) = (2+44t, 2+14t, 7-20t)

If you like, you can remove a common factor of 2 from the coefficients of t.

(x, y, z) = (2+22t, 2+7t, 7-10t)

5 0

3 years ago