Answer:
I dont know sorry
Step-by-step explanation:
Answer:
Domain of the inverse is (5, infinity)
Range of the inverse is (-infinity , -2)
Step-by-step explanation:
To find the domain of the inverse function, we have that;
The range of the function becomes the domain of the inverse
The domain of the function becomes the range of its inverse
so the domain is (5, infinity)
range is (-infinity, -2)
1) surface of a rectangular prism=2(length x width)+2(length x height)+2(width x height)
Therefore:
148 cm²=2(5 cm x 4 cm)+2(5 cm x h)+2(4 cm x h)=
148 cm²=40 cm²+10 cm h+8 cm h
18 cm h=148 cm²-40 cm²
18 cm h=108 cm²
h=108 cm² / 18 cm=6 cm.
answer: height=6 cm
2)
Volume of a rectangular prism= length x width x height
therefore:
34 cm³=(1.7 cm)(0.5 cm) h
0.85 cm² h=34 cm³
h=34 cm³/0.85 cm²
h=40 cm.
answer: height=40 cm
3)
volume of a cylinder: πr²h
therefore.
118.79 ft³=πr²(5 ft)
r=√(118.79 ft³/5π ft)≈2.75 ft
answer: radius=2.75 ft
4)
Surface area of the pyramid with square base=4(A side)+A base
A side=(1/2)(8ft)(12 ft)=48 ft²
A base=(8 ft)(8 ft)=64 ft²
surface area=4(48 ft²)+64 ft²=256 ft²
Answer: the surface area of this pyramid would be 256 ft².
5)
surface of a cone=πrs+πr²
therefore:
radius=diameter/2=6.2 ft/2=3.1 ft
63.3 ft²=π(3.1 ft) s+π(3.1 ft)²
3.1π ft s=33.109 ft²
s=33.109 ft² /3.1π ft
s≈3.4 ft
Answer: the slant height would be 3.4 ft.
6)
volume of a square pyramid=(area of base x heigth)/3
therefore:
area of base=(6 ft)(6 ft)=36 ft²
126.97 ft³=36 ft² h /3
h=126.97 ft³/12 ft²=10.58 ft
answer: the height would be 10.58 ft.
7)
volume of a cone =(base x height)/3
base of a cone=πr²
therefore:
199.23 cm³=πr²(9 cm)/3
r=√(199.23 cm³ / 3π cm)≈4.6 cm
answer: the radius would be 4.6 cm.
Answer:
<h2>The answer is 1</h2>
Step-by-step explanation:
The slope of the line between two points can be found by using the formula
where
(x1 , y1) and (x2 , y2) are the points
From the question is points are
(8, 1) and (-1, 10)
Substitute the values into the above formula and solve for the slope
That's
We have the final answer as
<h3>1 </h3>
Hope this helps you
