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svp [43]
3 years ago
6

3×3×5×5 as powers??​

Mathematics
2 answers:
Feliz [49]3 years ago
6 0

Answer:

= 225

Step-by-step explanation:

3 × 3 × 5 × 5

= 3² x 5²

= 9 x 25

<em><u>= 225</u></em>

eimsori [14]3 years ago
6 0

Answer:

= 225

Step-by-step explanation:

3 × 3 × 5 × 5

= 3² x 5²

= 9 x 25

= 225

Hope this helps :)

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3/8 + 1/8 add or subtract simplified
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3/8 + 1/8 = 4/8 = 1/2
The simplified answer is 1/2.
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The equation y=8.50x + 33 can be used to estimate the monthly premium y of a $250,000 term life insurance policy for a male smok
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Answer:

$101

Step-by-step explanation:

Given

y = 8.50x + 33

x \to age above 40

Required

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x = 48 - 40

x = 8

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y = 8.50x + 33

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What is the y-value in the solution to this system of linear equations?
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Step-by-step explanation:

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3 years ago
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RHOMBUS The diagonals of rhombus ABCD intersect at E. Given that
schepotkina [342]

Answer:

<u>Question 11:</u>

\angle DAC = 53^\circ

\angle AED = 90^\circ

\angle ADC = 74

DB = 16

AE = 6.03

AC = 12.06

<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

So:

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>

AC and BD are perpendicular lines, and they are diagonals

4 0
3 years ago
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