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just olya [345]
3 years ago
9

Which of the following statements says that a number is between -5 and 5

Mathematics
2 answers:
ollegr [7]3 years ago
5 0
I think it’s |x| < 5
Serhud [2]3 years ago
4 0

Answer:

i highly reccomend its on android and iphone you can use ur camera

You might be interested in
Write an equivalent expression to: 5(2x + 3)
zalisa [80]

Answer:

10x + 15

Step-by-step explanation:

Multiplying by 5 gives the aforementioned answer.

7 0
3 years ago
You are build a water tower. The scale factor is 1 : 12. The model depth is 32 centimeters. What is the actual depth in meters?
sattari [20]
Scale Factor = 1 : 12

It means, for every 12 centimeters, it is 1 centimeter in the model

The depth of Tower in model = 32 cm
In real, It would be: 32 * 12 = 384 cm

We know, 1 m = 100 cm
So, 384 cm = 384/100 = 3.84 m

In short, Your Answer would be: 3.84 meter

Hope this helps!
7 0
3 years ago
Circle O has a circumference of 367 cm.
olya-2409 [2.1K]

Answer:

  none of the above

Step-by-step explanation:

The applicable relationship is ...

  C = 2πr

Solving for r, we get

  r = C/(2π) ≈58. (367 cm)/(2·3.14159) ≈ 58.41 cm . . . . no matching choice

_____

When the problem does not include a correct answer choice, I usually suggest you ask your teacher to show you the working of it.

7 0
3 years ago
Read 2 more answers
Please helpp with my math hw I’ll give brainlest
-Dominant- [34]

Answer:

20 percent

40 percent

20 percent

20(.008) percent

Step-by-step explanation:

6) 54 - 45 = 9

45 / 9 = 5

100 / 5 = 20

20 percent

7) 60 - 36 = 24

60 / 24 = 2.5

100 / 2.5 = 40

40 percent

8) 30 - 24 = 6

30 / 6 = 5

100 / 5 = 20

20 percent

9) 24.99 - 19.99 = 5

24.99 / 5 = 4.998

100 / 4.998 approx. = 20.008

20.008 percent (or rounded 20 percent)

4 0
2 years ago
A cylindrical rod of steel (E = 207 GPa, 30 * 106 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a l
sineoko [7]

Answer:9.477 mm

Step-by-step explanation:

Given

E=207 GPa

Yield Strength(s_{yt}) 310 MPa

load (P)=11,100 N

length of rod(L)=500 mm

\Delta L=0.38 mm

we know \Delta L is given by

\Delta =\frac{PL}{AE}

where A= cross-section

A=\frac{11100\times 500}{0.38\times 207\times 10^9}

A=70.556\times 10^{-6} m^2

A=\frac{\pi d^2}{2}=70.556\times 10^{-6} m^2

d=9.477\times 10^{-3} m

d=9.477 mm

4 0
4 years ago
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