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Lynna [10]
2 years ago
14

5.1 m 7.2 m 1.2 m 2.1 m 4.5 m​

Mathematics
2 answers:
Arisa [49]2 years ago
5 0

What's the question? there isn't enough info to help.

masya89 [10]2 years ago
3 0

Answer:

there isn't much information to help. pls try. check question

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Help Me, Will give 100 points.
Bad White [126]

Answer:

baka

Step-by-step explanation:

6 0
2 years ago
2) Write the first five terms of the arithmetic sequence, if:<br> a3=8<br> a11=32
padilas [110]

Answer:

a

Step-by-step explanation:

the five terms for arithmetic is that

5 0
2 years ago
What is the shape of the the cross section<br> PLEASE HELP
Masteriza [31]

Answer: The shape of the cross section will be triangular.

Step-by-step explanation:

The cross section is the shape the shows when you cut a straight line. no matter where you cut it, you will still get a triangle.

4 0
3 years ago
What is the approximate volume of a cylinder with a diameter of 12 meters and a height of 7 meters? Use 3.14 for pi.
kkurt [141]
V=\pi r^2 h
r - radius, h - height
The diameter is twice the radius.

d=12 \ m \\&#10;r=\frac{12}{2} \ m = 6 \ m \\&#10;h=7 \ m \\ \\&#10;V=\pi \times 6^2 \times 7=\pi \times 36 \times 7=252 \pi \approx 252 \times 3.14=791.28 \approx 791.3

The answer is c. 791.3 m³.
3 0
3 years ago
Consider the inverse function. Which conclusions can be drawn about f(x) = x2 + 2? Select three options. f(x) has a limited rang
PolarNik [594]

Answer:

f(x) has a limited range

f(x) has a maximum at the point (0, 2)

f(x) has a y-intercept at the point (0, 2).

Step-by-step explanation:

Given the function;

f(x) = x^2+2

The domain is the value of the input variables for which the function will exist. According to the expression given, the function exists on all real values of x. The same goes with range which deals with the output values. It also exists on all real values from 2 and above.

Hence f(x) have a limited range (since values less than 2 are not included compare to domain that exists on all real values) and does not have a restricted domain.

For the x intercept, x intercept occur at y = 0

substitute y = 0 into the function and get y

if y = f(x)

y = x^2+2

0 = x^2 + 2

x^2 = -2

x = 2i

Hence  f(x) does not have an x-intercept of (2, 0)

For the y intercept, y intercept occur at x = 0

substitute x = 0 into the function and get y

if y = f(x)

y = x^2+2

y = 0^2 + 2

y = 2

Hence  f(x) has a y-intercept at point (0, 2)

f(x) is at maximum if d(fx))/dx = 0

d(fx))/dx  = 2x

since  d(fx))/dx  = 0

0 = 2x

x = 0

substitute x = 0 into the function

f(x) = x^2 + 2

y = 0^2+2

y = 2

Hence f(x) has a maximum at the point (0, 2)

5 0
2 years ago
Read 2 more answers
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