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Firdavs [7]
3 years ago
8

PLEASE HELP. WILL GIVE BRAINLIEST!!!!!!!!!

Mathematics
2 answers:
Alex777 [14]3 years ago
7 0

Answer:

A

Step-by-step explanation:

dalvyx [7]3 years ago
6 0

Answer:I think it is a

Step-by-step explanation:

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How many dimes are equal to $9? (10 dimes = $1.00)
Ivanshal [37]
90 dimes. Hope I helped
3 0
3 years ago
Read 2 more answers
Mario and his family went to the county fair.They bought 2 adult passes.what was the total cost for the family?
joja [24]

<u>Answer:</u>

The total cost for the family was $16

<u>Step-by-step explanation:</u>

Given:

Number of adult tickets = 2

Number of children ticket = 3

To Find:

The total cost of the family = ?

Solution:

The total cost of the family   =  cost for buying 2 adult tickets + cost of buying 3 children tickets

=> 2 \times 5 + 3 \times 2

=> 10 + 6

=> 16

7 0
4 years ago
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
0.04(y - 9) +0.16y = 0.12y-0.5<br> y=<br> (Type an integer or a decimal.)
Vladimir [108]
The answer is going to be y=-1.75 hope this helps!!

3 0
3 years ago
BAD is congruent to BCD by <br> the
scoundrel [369]
AB is congruent to BC
6 0
4 years ago
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