The base is 3+(x), altitude is x so substitute. Now we know the area of a triangle is base X height X 1/2. Substitute again! 1/2 (3+x)(x)=35. Multiply both sides by 2 to cancel out the 1/2. Now you have (x)(x+3)=70 and you have to foil out the left side x^2+3x=70. Subtract 70 on both sides x^2+3x-70=0. Find two numbers that multiply to -70 and add to 3. Solve (x+10)(x-7)=0. the x value is 7. since you can't have negative length values. Substitute 7 into 3+x for the base so you know the base is 10 and the height is 7.
<span>Distance from point (4,1,-1) to plane x-2y+2z+3 = 0 is 1.
The formula for the distance from a point to a plane is
d = abs(Ax + By + Cz + D) / sqrt(A^2 + B^2 + C^2)
where
d = distance
(x,y,z) = coordinates of point
A,B,C,D = coefficients of equation for the plane
For this problem, the values for A,B,C,and D are A=1, B=-2, C=2, and D=3. So substituting the known values into the equation, we get:
d = abs(1*x - 2*y + 2*z + 3) / sqrt(1^2 + (-2)^2 + 2^2)
d = abs(1*4 - 2*1 + 2*(-1) + 3) / sqrt(1 + 4 + 4)
d = abs(4 - 2 - 2 + 3) / sqrt(9)
d = abs(3) / 3
d = 1</span>
Answer:
I have
Step-by-step explanation:
No idea sorry
Answer:
3x^2(25x^2−xy+5y^2)
Step-by-step explanation:
I copied it into MathPapa and it is always correct. It gave me that