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Tanzania [10]
3 years ago
6

The real numbers whose decimals do not end and do not repeat are (irrational numbers / rational numbers).

Mathematics
2 answers:
Zigmanuir [339]3 years ago
6 0

Answer:

Irrational numbers

Step-by-step explanation:

Rational numbers are integers. They don't have repeating decimals.

An irrational number is not an integer, because irrational number are <em>"not rational"</em> and all integers can be represented as rational, so irrational numbers decimals repeat.

pantera1 [17]3 years ago
3 0

9514 1404 393

Answer:

  irrational numbers

Step-by-step explanation:

The real numbers whose decimals do not end and do not repeat are irrational numbers.

__

If a decimal number ends or repeats, it can be represented by the <em>ratio</em> of two integers. That is the essence of a <em>ratio</em>nal number.

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40.) Decompose 7/8 into the sum of unit fractions.<br>​
Paraphin [41]

Answer:

7/8 = 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8

7/8 = 1/8 + 6/8

7/8 = 4/8 + 3/8

7/8 = 5/8 + 2/8

Step-by-step explanation:

Hope it helps!

4 0
4 years ago
2x-4<br> 30<br> 15x - 30<br> PAP
Alex787 [66]

Answer:

\frac{2}{25}

Step-by-step explanation:

Given expression,

\frac{\frac{2x-4}{30} }{\frac{15x-30}{18} }

=>\frac{\frac{2(x-2)}{30} }{\frac{15(x-2)}{18} }

We see that (x-2) is common in both numerator and denominator

hence canceling out we get:-

\frac{2*18}{30*15}=\frac{2}{25}

Answer=\frac{2}{25}

4 0
3 years ago
The line L passes through the points (5, -5) and (1, 2).<br> Find the gradient of line L.
Free_Kalibri [48]

gradient of L = -7/4

Explanation:

\sf gradient = \dfrac{y_2-y_1}{x_2-x_1}

<u>Insert the following values.</u>

\rightarrow \sf \dfrac{2-(-5)}{1-5}

\rightarrow \sf \dfrac{7}{-4}

\rightarrow \sf -\dfrac{7}{4}

7 0
2 years ago
Read 2 more answers
Solve for k.<br> 11<br> 8<br> k<br> 4<br> -
dusya [7]

Answer:

Step-by-step explanation:

11×8×k -4= 0

88k. = 4

K. = 1/22

6 0
3 years ago
Read 2 more answers
On the basis of the graph, answers the following questions.
Pepsi [2]

a) The factors have the form:

(x-x_1)

where x1 is a zero of the function. A zero is a point at which the graph intercepts the x-axis. From the graph, the zeros are:

-6, -4, 2, and 3

Therefore, the factors are:

(x + 6)

(x + 4)

(x - 2)

(x - 3)

b) Multiplying all these factors we get a polynomial, p(x), with the zeros of the graph. That is:

p(x) = (x + 6)(x + 4)(x - 2)(x - 3)

c) Yes, it is possible to find other polynomials with the same zeros. To do that we have to multiply p(x) by a constant. For example, multiplying by 2:

f(x) = 2(x + 6)(x + 4)(x - 2)(x - 3)

and f(x) has the same zeros as p(x)

d) Every polynomial obtained in the previous way, multiplying p(x) by a constant, will have a different graph. In conclusion, it is not possible to find other polynomials with the same zeros and the same graph.

6 0
1 year ago
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