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Andreyy89
3 years ago
14

Is anyone able to help me?

Mathematics
1 answer:
bogdanovich [222]3 years ago
7 0

Answer:

The volume of this triangular prism is 20cm^{3}

Step-by-step explanation:

Just like it is shown in the picture we can use the formula for the volume of a triangular prism.

V = \frac{1}{2}bhl

In our case b = 5cm, h = 2 cm, l = 4cm. Now we just plug in the values and get...

V = \frac{1}{2}bhl\\\\

V = \frac{1}{2}(5)(2)(4)

V = 20cm^{3}

Therefore the volume of the triangular prism is 20cm^{3}.

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Round 0.00543998 to 2 significant figures
Ghella [55]

Answer:

0.0053

Step-by-step explanation:

Look at the 2 significant figures which are non-zero digits.

The two figures are 5 and 4, the digit after 4 is 3, which is lower than 5, so we don’t have to add +1 to the 4.

5 0
3 years ago
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What is the midpoint of the line segment graphed below?
JulsSmile [24]

Answer:

a) (4,7/2)

Step-by-step explanation:

Midpoint Formula:

[(-1+9)/2], [(2+5)/2]

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3 years ago
Which of the following does not represent a function?
ankoles [38]
<h3>Answer:</h3>

graph of a parabola symmetric about the x axis with x intercept at negative 5

<h3>Explanation:</h3>

A parabola that opens horizontally does not pass the vertical line test. You can draw a verticla line that intersects the curve in 2 places. A function must be single-valued, so can only have one y-value for any x-value.

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3 years ago
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Someone start a convo in the comments or try and message me idk how
ivanzaharov [21]

Answer:

Brainly is a platform that is used to help students in need. There are many social networks that offer socializing with many people from every country in the world.

I suggest you install any of those apps, and try find someone to talk to.

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Hope I helped.

6 0
3 years ago
What are the orders of 3,7,9,11,13,17 and 19(mod20)?does 20 have primitive roots?
bezimeni [28]
3\equiv3\mod{20}
3^2\equiv9\mod{20}
3^3\equiv27\equiv7\mod{20}
3^4\equiv3\cdot3^3\equiv3\cdot7\equiv21\equiv1\mod{20}

7\equiv7\mod{20}
7^2\equiv49\equiv9\mod{20}
7^3\equiv7\cdot7^2\equiv63\equiv3\mod{20}
7^4\equiv7\cdot7^3\equiv21\equiv1\mod{20}

9\equiv9\mod{20}
9^2\equiv3^4\equiv1\mod{20}

11\equiv11\mod{20}
11^2\equiv121\equiv1\mod{20}

13\equiv-7\equiv13\mod{20}
13^2\equiv169\equiv9\mod{20}
13^3\equiv13\cdot13^2\equiv(-7)9\equiv-63\equiv-3\mod{20}
13^4\equiv13\cdot13^3\equiv(-7)(-3)\equiv21\equiv1\mod{20}

17\equiv-3\equiv17\mod{20}
17^2\equiv(-3)^2\equiv9\mod{20}
17^3\equiv(-3)^3\equiv-27\equiv3\mod{20}
17^4\equiv(-3)^4\equiv81\equiv1\mod{20}

19\equiv-1\equiv19\mod{20}
19^2\equiv19(-1)\equiv-19\equiv1\mod{20}

Generally speaking, a number x coprime to n will be a primitive root of n if we have x^n\equiv x\mod{n}, or x^{n-1}\equiv1\mod{n}. In other words, if x is of order n-1 modulo n, then x is a primitive root of n.

Since none of these numbers has order 19, it follows that 20 does not have any primitive roots.
6 0
4 years ago
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