7x+5y=-24 (1)
4x+y=42 (2)
multiply equation (2) by 5 to get
20x+5y=210 (3)
then calculate (3)-(2) which gives you
13x=234 hence x=18
then substitute for x in either equation to get y=-30
Answer:
1 : equal in force, amount, or value also : equal in area or volume but not superposable a square equivalent to a triangle. 2a : like in signification or import. b : having logical equivalence equivalent statements. 3 : corresponding or virtually identical especially in effect or function.
Step-by-step explanation:
Answer:
<h2>
∠PQT = 72°</h2>
Step-by-step explanation:
According to the diagram shown, ∠OPQ = ∠OQP = 18°. If PQT is a tangent to the circle, it can be inferred that line OQ is perpendicular to line QT. Ths shows that ∠OQT = 90°.
Also from the diagram, ∠OQP + ∠PQT = ∠OQT;
∠PQT = ∠OQT - ∠OQP
Given ∠OQP = 18° and ∠OQT = 90°
∠PQT = 90°-18°
∠PQT = 72°
Answer:
Step-by-step explanation:
Question
Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work.
perimeter of a triangle = AB+AC+BC
Using the distance formula
AB = sqrt(-2-5)²+(2-2)²
AB = sqrt(-7)²
AB =sqrt(49)
AB =7
BC = sqrt(-2+2)²+(2-5)²
BC = sqrt(0+3²)
BC =sqrt(9)
BC =3
AC= sqrt(-2-5)²+(2-5)²
AC= sqrt(-7)²+3²
AC =sqrt(49+9)
AC =sqrt58
Perimeter = 10+sqrt58
Answer:
Eq: (x+a/2)²+(y+1)²=(a²-8)/4
Center: O(-a/2, -1)
Radius: r=0.5×sqrt(a²-8)
Mandatory: a>2×sqrt(2)
Step-by-step explanation:
The circle with center in O(xo,yo) and radius r has the equation:
(x-xo)²+(y-yo)²=r²
We have:
x²+y²+ax+2y+3=0
But: x²+ax=x²+2(a/2)x+a²/4-a²/4= (x+a/2)²-a²/4
And
y²+2y+3=y²+2y+1+2=(y+1)²+2
Replacing, we get:
(x+a/2)²-a²/4+(y+1)²+2=0
(x+a/2)²+(y+1)²=a²/4-2=(a²-8)/4
By visual inspection we note that:
- center of circle: O(-a/2, -1)
- radius: r=sqrt((a²-8)/4)=0.5×sqrt(a²-8). This means a²>8 or a>2×sqrt(2)
