PQ = 9
QR = 28
Note, there is a Q in each line
Combine the two lines: PQ + QR = PR
plug in the numbers to corresponding variables
9 + 28 = PR
PR = 37
hope this helps
Answer:
124.67cm³
Step-by-step explanation:
==>Given:
Dimensions of current can:
Height (h) = 12cm
Diameter = 6cm (radius = 3cm)
Volume of current can (V1) = 339.12 cm³
Dimensions of future cans to be increased multiple of 1.11:
height = 12cm × 1.11 = 13.32 cm
radius = 3cm × 1.11 = 3.33 cm
Volume of future can (V2) = πr²h = 3.14*3.33²*13.32
= 3.14*11.0889*13.32 = 463.791025
V2 ≈ 463.79 cm³
==>Find how much more would new cans hold = Volume of new can - volume of current can
= 463.79 cm³ - 339.12 cm³
= 124.67cm³
Answer:
Yes he has enough fence to go round the flower bed.
Step-by-step explanation:
1 yard = 3 feet
25 feet = 25/3 = 8.33 yards
Yes he has enough fence to go round the flower bed.
Answer:
- hexahedron: triangle or quadrilateral or pentagon
- icosahedron: quadrilateral or pentagon
Step-by-step explanation:
<u>Hexahedron</u>
A hexahedron has 6 faces. A <em>regular</em> hexahedron is a cube. 3 square faces meet at each vertex.
If the hexahedron is not regular, depending on how those faces are arranged, a slice near a vertex may intersect 3, 4, or 5 faces. The first attachment shows 3- and 4-edges meeting at a vertex. If those two vertices were merged, then there would be 5 edges meeting at the vertex of the resulting pentagonal pyramid.
A slice near a vertex may create a triangle, quadrilateral, or pentagon.
<u>Icosahedron</u>
An icosahedron has 20 faces. The faces of a <em>regular</em> icosahedron are all equilateral triangles. 5 triangles meet at each vertex.
If the icosahedron is not regular, depending on how the faces are arranged, a slice near the vertex may intersect from 3 to 19 faces.
A slice near a vertex may create a polygon of 3 to 19 sides..
Answer:
Let t be the number of ten dollar bills.
o×6+5=t
If you need it without t, then it's:
o×6+5
Step-by-step explanation:
Let t be the number of ten dollar bills.
o×6+5=t
We get this because the question said that there were more ten-dollar bills than 6 times the number of one-dollar bills.