Answer:
We need a non-included side of one triangle
Step-by-step explanation:
By means of the AAS postulate.
The Angle-Angle-Side postulate (AAS) tells us that if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the two triangles are congruent.
Answer:
<h2>One</h2>
Step-by-step explanation:
Given the value 78247 a s a stable number because at least one of its digits has the same value as its position in the number. The 4th number in the value is 4, this makes the number a stable number.
The following are the 3-digits stable numbers that appears in 78247
The first number is 824. This digits are stable numbers because 2 as a number is situated in the same place as the number (2nd position).
Hence, there are only 1 stable 3-digit numbers in the value 78247 since only a value exists as 2 in the value and there is no 1 and 3 in the value.
Hey
So my brother posted this on Yahoo
Draw a line from the center of the circle to one of the ends of the chord (water surface) and another to the point at greatest depth. A right-angled triangle is formed. Length of side to the water-surface is 5 cm, the hypot is 7 cm.
<span>What you do now is the following: </span>
<span>Calculate the angle θ in the corner of the right-angled triangle by: cos θ = 5/7 ⇒ θ = cos ˉ¹ (5/7) </span>
<span>So θ is approx 44.4°, so the angle subtended at the center of the circle by the water surface is roughly 88.8° </span>
<span>The area shaded will then be the area of the sector minus the area of the triangle above the water in your diagram. </span>
<span>Shaded area ≃ 88.8/360*area of circle - ½*7*7*sin88.8° </span>
<span>= 88.8/360*π*7² - 24.5*sin 88.8° </span>
<span>≃ 13.5 cm² </span>
<span>(using area of ∆ = ½.a.b.sin C for the triangle) </span>
<span>b) </span>
<span>volume of water = cross-sectional area * length </span>
<span>≃ 13.5 * 30 cm³ </span>
<span>≃ 404 cm³</span>
Hoped it Helped
Answer:
x is less-than-or-equal-to 2.25 (x ≤ 2.25)
Step-by-step explanation:
We can write down the inequality that represents the weight Li can add without going over the 50 pound limit:
47.75 + x ≤ 50
If we solve for x we have:
47.75 + x ≤ 50
x ≤ 50 - 47.75
x ≤ 2.25
Therefore, the weight Li can add to the suitcase is less-than-or-equal-to 2.25
