Answer:
3. ASA
4. Corresponding sides of congruent triangles are congruent
Step-by-step explanation:
I marked this in the attached image. ∠B ≅∠E, ∠ACE ≅ ∠DCE, and the side between them BC ≅ EC, so the the triangles ΔACB and ΔDCE are congruent by ASA.
If the triangles are congruent, then all corresponding angles and sides are also congruent.
3
x
+
2
y
>
24
3
x
+
2
y
>
24
Solve for
y
y
.
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y
>
−
3
x
2
+
12
y
>
-
3
x
2
+
12
Use the slope-intercept form to find the slope and y-intercept.
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Slope:
−
3
2
-
3
2
Y-Intercept:
12
12
Graph a dashed line, then shade the area above the boundary line since
y
y
is greater than
−
3
x
2
+
12
-
3
x
2
+
12
.
y
>
−
3
x
2
+
12
y
>
-
3
x
2
+
12
Common factor that can be multiplied to get that product
0 to 100 miles per hour<span> is the answer so now can i get a hug</span>
Answer:
v = 1/(1+i)
PV(T) = x(v + v^2 + ... + v^n) = x(1 - v^n)/i = 493
PV(G) = 3x[v + v^2 + ... + v^(2n)] = 3x[1 - v^(2n)]/i = 2748
PV(G)/PV(T) = 2748/493
{3x[1 - v^(2n)]/i}/{x(1 - v^n)/i} = 2748/493
3[1-v^(2n)]/(1-v^n) = 2748/493
Since v^(2n) = (v^n)^2 then 1 - v^(2n) = (1 - v^n)(1 + v^n)
3(1 + v^n) = 2748/493
1 + v^n = 2748/1479
v^n = 1269/1479 ~ 0.858
Step-by-step explanation:
