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3 years ago

Which numbers have a digit in the ones place that is 1/10 the value of the digit in the tens place

Mathematics

1 answer:

liq [111]3 years ago

7 0

Answer:

4,099 and 5,011

Step-by-step explanation:

This problem can be solved by taking options one by one.

Option (1) : 4,099

Digit in ones place = 9

The value of the digit in tens place = 90

$\dfrac{9}{90}=\dfrac{1}{10}$ . It is correct.

Option (2) : 4,110

Digit in one places = 0

The value of the digit in tens place = 10

It is incorrect.

Option (3) : 5,909

Digit in one places = 9

The value of the digit in tens place = 0

It is again incorrect.

Option (4) : 5,011

Digit in one places = 1

The value of the digit in tens place = 10

$\dfrac{1}{10}$ . It is correct.

Hence, in option (a) and (d), the he ones place is 1/10 the value of the digit in the tens place.

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I am confused about how to do this problem and I don't think it should be hard, but I just don't know how to approach it, so if

solniwko [45]

Answer:

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) (0.875, 0.676)

d) (0, 1.235)

Step-by-step explanation:

Set each term in the numerator and denominator equal to 0 and find r.

In the numerator:

r = 7/8, 5/9, or 6

In the denominator:

r = 9/5, 7/8, or -3

Zeros in the numerator that aren't in the denominator are r-intercepts.

Zeros in the denominator that aren't in the numerator are vertical asymptotes.

Zeros in both the numerator and the denominator are holes.

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) Evaluate m(r) at r = 7/8. To do that, first divide out the common term (-8r + 7) from the numerator and denominator.

m(r) = (-9r+5)² (r−6)² / ( (-5r+9)² (r+3)² )

m(⅞) = (-9×⅞+5)² (⅞−6)² / ( (-5×⅞+9)² (⅞+3)² )

m(⅞) = (-23/8)² (-41/8)² / ( (37/8)² (31/8)² )

m(⅞) = (-23)² (-41)² / ( (37)² (31)² )

m(⅞) = 0.676

The hole is at (0.875, 0.676).

d) Evaluate m(r) at r = 0.

m(0) = (-9×0+5)² (0−6)² / ( (-5×0+9)² (0+3)² )

m(0) = (5)² (-6)² / ( (9)² (3)² )

m(0) = 1.235

The m(r)-intercept is (0, 1.235).

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3 years ago