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3 years ago

13

Three times a larger number is 30 more than 5 times a smaller number. The sum of the larger number and 5 times the smaller numbe

r is 50

1 answer:

kicyunya [14]3 years ago

7 0

Answer: Smaller number = 6

Larger number = 20

Step-by-step explanation:

Let the smaller number be x.

Let the larger number be y.

Three times a larger number is 30 more than 5 times a smaller number. This will be:

3 × y = (5 × x) + 30

3y = 5x + 30 ...... i

The sum of the larger number and 5 times the smaller number is 50. This will be:

y + 5x = 50 ...... ii

From equation ii

y + 5x = 50.

y = 50 - 5x ...... iii

Put the value of y = 50 - 5x into equation i

3y = 5x + 30

3(50 - 5x) = 5x + 30

150 - 15x = 5x + 30

150 - 30 = 5x + 15x

120 = 20x

x = 120/20

x = 6

The smaller number is 6

Note that y + 5x = 50

y + 5(6) = 50

y + 30 = 50

y = 50 - 30

y = 20

The bigger number is 20

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Answer:

<u>Question 11:</u>

$\angle DAC = 53^\circ$

$\angle AED = 90^\circ$

$\angle ADC = 74$

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<u>Question 12:</u>

$\triangle ABD$ , $\triangle BAC$ , $\triangle CDA$ and $\triangle DAB$

<u>Question 13: </u>

AC and BD are perpendicular lines, and they are diagonals

Step-by-step explanation:

<u>Question 11</u>

Given

$\angle BAC = 53^\circ$

$DE = 8$

See attachment for Rhombus

Required

Determine the indicated sides

Solving (a): $\angle DAC$

Diagonal CA divides $\angle DAB$ into 2 equal angles

i.e

$\angle DAC = \angle BAC$

So:

$\angle DAC = 53^\circ$

Solving (b): $\angle AED$

The angles at E is 90 degrees because diagonals AC and BD meet at a perpendicular.

So:

$\angle AED = 90^\circ$

Solving (c): $\angle ADC$

First, we calculate $\angle ADE$ , considering $\triangle ADE$ :

$\angle ADE + \angle AED + \angle DAC = 180$

$\angle ADE + 90 + 53 = 180$

$\angle ADE + 143 = 180$

$\angle ADE = -143 + 180$

$\angle ADE = 37$

To calculate $\angle ADC$ , we have:

$\angle ADC = 2*\angle ADE$

$\angle ADC = 2* 37$

$\angle ADC = 74$

Solving (d): $DB$

From the rhombus

$DB = DE +EB$

Where

$DE =EB$

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$DB = 8 + 8$

$DB = 16$

Solving (e): $AE$

To do this we consider $\triangle ADE$

Using the tan formula

$tan(\angle ADE) = \frac{AE}{DE}$

$\angle ADE = 37$ and $DE = 8$

So:

$\tan(37) = \frac{AE}{8}$

$AE = 8 * \tan(37)$

$AE = 6.03$

Solving (f): $AC$

This is calculated as:

$AC = AE + EC$

Where

$AE = EC$

$AC = 6.03 +6.03$

$AC = 12.06$

<u>Question 12: Isosceles Triangle</u>

In the rhombus, all 4 sides are equal;

So, the isosceles triangle are:

$\triangle ABD$ , $\triangle BAC$ , $\triangle CDA$ and $\triangle DAB$

<u>Question 13: </u>

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olya-2409 [2.1K]

Answer:

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Step-by-step explanation:

As with all formulas, put the numbers in place of the corresponding variables, and do the arithmetic.

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The first 5 terms of the sequence are 3, 8, 14, 30, 58.

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