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2 years ago

Which statement cannot be justified by one of the properties of real numbers

Mathematics

1 answer:

nexus9112 [7]2 years ago

6 0

The answer would be B because it demonstrates the fact that if a is decreased by the quotient of b and c then the second expression wouldn't result into that same answer due to the different arrangements of the signs.

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An astronaut who weighs 85 kilograms on earth weighs 14.2 kilograms on the moon. How much would a person weigh on the moon if th

STatiana [176]

Answer:

Weight on the moon=x=15.87 kilograms

Step-by-step explanation:

This can be expressed as;

Weight on earth=Constant×Weight on moon

where;

Weight on moon=14.2 kilograms

Constant=k

Weight on Earth=85 kilograms

Replacing in the expression above;

85=14.2×k

k=(85/14.2)=5.986

If the person weighs 95 kilograms on Earth;

Weight on the earth=Constant×Weight on moon

where;

Weight on the moon=x

Constant=5.986

Weight on the earth=95 kilograms

Replacing;

95=5.986×x

x=95/5.986=15.87

Weight on the moon=x=15.87 kilograms

4 0

2 years ago

5c) The scale of a map says that 4 cm represents 5 km. What is the actual

Maslowich

● Answer:

6.25 km

● Step-by-step explanation:

Hi there !

4 cm ........... 5 km

5 cm ........... x km

x = 5cm×5km/4cm

= 25km/4

= 6.25 km

Good luck !

5 0

3 years ago

Two lines or equations are described in each part. Decide whether each system has one solution, no solution, or infinitely many

Vedmedyk [2.9K]

It has no solution bc there or no point and because its not the same line

7 0

3 years ago

A garden box has a perimeter of 18½ feet. If the length is 5 feet, what is the area of the garden box?

Travka [436]

Answer:

$area=\frac{85}{4}feet^{2}$

Step-by-step explanation:

1. Approach

Since it is given that the garden box is a rectangle, then the opposite sides are congruent. One can use this to their advantage, by setting up an equation that enables them to solve for the width of the rectangle. After doing so, one will multiply the width by the given length and solve for the area.

2. Solve for the width

It is given that the garden box is a rectangle. As per its definition, opposite sides in a rectangle are congruent. The problem gives the length and the perimeter of the rectangle, therefore, one can set up an equation and solve for the width.

$perimeter=18\frac{1}{2}\\\\length=5$

$2(length+width)=perimeter$

Substitute,

$2(5+width)=18\frac{1}{2}$

Conver the mixed number to an improper fraction. This can be done by multiplying the "number" part of the mixed number by the denominator of the fraction. Then add the result to the numerator.

$2(5+width)=\frac{37}{2}$

Inverse operations,

$2(5+width)=\frac{37}{2}\\/2\\\\5+width=\frac{37}{4}\\-5\\\\width=\frac{17}{4}$

3. Solve for the area

Now that one has solved for the width of the box, one must solve for the area. This can be done by multiplying the length by the width. Since the width is a fraction, one must remember, that when multiplying an integer by a fraction, one will multiply the integer by the numerator (the top of the fraction), and then simplify by reducing the fraction, if possible. Reducing the fraction is when one divides both the numerator and the denominator by the GCF (Greatest Common Factor).

$length=5\\\\width = \frac{17}{4}$