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3 years ago

Which product is less than 99 (look at pic) please answer

Mathematics

2 answers:

marin [14]3 years ago

5 0

Answer: 10 / 11 × 99

Step-by-step explanation:

The first one is; 16/15 × 99 = 1584/15 = 105.6

Its greater than 99

The second one is; 12/7 × 99 = 1188/7 = 169.71

The value is greater than 99

The fourth one is; 20/20 × 99 = 1980/20 = 99

The value is equal 99

The third one is; 10/11 × 99 = 990/11 =90

The value is less than 99

Therefore the product of 10/11 ×99 is less than 99

ahrayia [7]3 years ago

3 0

The third one because the fraction is less than one

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Answer:

x = 1/5, y = 6

Step-by-step explanation:

5x + y = 7

y = 7 - 5x

Set equations equal to each other to eliminate y

7 - 5x = 20x + 2

7 = 25x + 2

5 = 25x

x = 1/5

y = 7 - 5(1/5)

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Ships a and b or 1425 feet apart and detect a submarine below them

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Answer:

$AX = 1084.20$

$BX = 1270.69$

Step-by-step explanation:

See attachment for complete question

Let the position of the submarine be represented with X

Given

$AB = 1425$

$\angle A = 59^{\circ}$

$\angle B = 47^{\circ}$

First, we calculate angle at X.

$\angle X + \angle A + \angle B = 180$

$\angle X + 59^{\circ} + 47^{\circ}= 180^{\circ}$

$\angle X = 180^{\circ} -59^{\circ} - 47^{\circ}$

$\angle X = 74^{\circ}$

Solving (a): Distance AX: The distance between ship A and the submarine

To do this, we apply sine formula which states

$\frac{a}{sin\ A} = \frac{b}{sin\ B} = \frac{c}{sin\ C}$

In this case:

$\frac{AB}{sin\ X} = \frac{AX}{sin\ B}$

Substitute values for AB, $\angle X$ and $\angle B$

$\frac{1425}{sin(74^{\circ})} = \frac{AX}{sin(47^{\circ})}$

Make AX the subject

$AX = \frac{1425}{sin(74^{\circ})} * sin(47^{\circ})$

$AX = \frac{1425}{0.9613} * 0.7314$

$AX = \frac{1425 * 0.7314}{0.9613}$

$AX = \frac{1042.245}{0.9613}$

$AX = 1084.20$

Solving (b): Distance BX: The distance between ship B and the submarine

To do this, we apply sine formula which states

In this case:

$\frac{AB}{sin\ X} = \frac{BX}{sin\ A}$

Substitute values for AB, $\angle X$ and $\angle A$

$\frac{1425}{sin(74^{\circ})} = \frac{BX}{sin(59^{\circ})}$

Make BX the subject

$BX = \frac{1425}{sin(74^{\circ})} * sin(59^{\circ})$

$BX = \frac{1425}{0.9613} * 0.8572$

$BX = \frac{1425* 0.8572}{0.9613}$

$BX = \frac{1221.51}{0.9613}$

$BX = 1270.69$

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Answer:

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Step-by-step explanation:

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