The gcf he can make is 131,313 packages
(8x - 35) solve for x
1 answer:
You might be interested in
Answer:
The lengths of MN is 22 units, MO is 26 units and JK is 10 units
Step-by-step explanation:
<em>A l</em><em>ine segment</em><em> joining the </em><em>mid-points of two sides</em><em> in a triangle is </em><em>parallel to the third side</em><em> and </em><em>equal to half its length</em>
In Δ MON
∵ J, K, and L are mid-points
∵ JL // MN and LK // MO
∴ L is the mid-point of ON
∴ J is the mid-point of MO
∴ K is the mid-point of MN
∵ J, L are the mid-points of MO and ON
∵ JL is opposite to MN
→ By using the rule above
∴ JL = MN
∵ JL = 11 units
∴ 11 = MN
→ Multiply both sides by 2
∴ 22 = MN
∴ MN = 22 units
∵ K, L are the mid-points of MN and ON
∵ KL is opposite to MO
→ By using the rule above
∴ KL = MO
∵ KL = 13 units
∴ 13 = MO
→ Multiply both sides by 2
∴ 26 = MO
∴ MO = 26 units
∵ J, K are the mid-points of MO and MN
∵ JK is opposite to ON
→ By using the rule above
∴ JK = ON
∵ ON =20 units
∴ JK = (20)
∴ JK = 10 units
∴ The lengths of MN are 22 units, MO is 26 units and JK is 10 units
Answer:
D
Step-by-step explanation:
We assume the rotation R is <em>counterclockwise</em> 60°.
__
The exponent on R is the number of times it is applied. That is, R² = R(R(figure)). So, the composition is equivalent to R^(2-4) = R^-2.
When the exponent of R is negative, it is essentially the inverse function. That is, applying the function R to the result will give the figure you started with. Equivalently, it is rotation in the other direction.
The point 120° clockwise from B is D.
The desired image point is D.