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juin [17]
3 years ago
14

Choose the equation of the line that is parallel to the y-axis.

Mathematics
1 answer:
marta [7]3 years ago
5 0

the answer to the problem is c

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What are the steps to this equation. 7 3/7 - 2 5/7
nignag [31]
Here is your answer:

7. 3×1=3
7×1=7


2. 5×1=5
7×1=7

=5 -2/7

Step one: write the equation

Step two: find the common denommarator (which is 7)

Step three: subtract 3-5= which equals to negative 2

Step four: subtract the whole numbers ( 7 and 2) which gives you five

Your answer is...

5 -2/7
3 0
4 years ago
Need help please!!!
forsale [732]
The answer is D
Explanation:5x^5-44x^3-9x=0
X(5x^2+1)(x-3)(x+3)=0
X(x+3)(x-3)=0
X=0,x=-3,x=3
3 0
3 years ago
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sweet [91]
Infinite Solutions :) Dhdhsjsjdnnd
8 0
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
Thanks for helping...
slava [35]

Answer:

16

Step-by-step explanation:

Subtracting the given expressions, that is

3b² - 8 - (b(b² + b - 7) ) ← simplify parenthesis

= 3b² - 8 - (b³ + b² - 7b) ← distribute parenthesis by - 1

= 3b² - 8 - b³ - b² + 7b ← collect like terms

= - b³ + 2b² + 7b - 8 ← substitute b = - 3

= - (- 3)³ + 2(- 3)² + 7(- 3) - 8

= - (- 27) + 2(9) - 21 - 8

= 27 + 18 - 21 - 8

= 16

7 0
3 years ago
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