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3 years ago

14

Carl is boarding a plane.

1 answer:

Reil [10]3 years ago

4 0

if this is an actual question could you give the whole thing?

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Evaluate x^2 + 2x + 3 <br> for x = 4

Mkey [24]

Steps to solve:

x^2 + 2x + 3 when x = 4

~Substitute

4^2 + 2(4) + 3

~Simplify

16 + 8 + 3

~Add

24 + 3

~Add

27

Best of Luck!

6 0

3 years ago

A farm has chickens and cows. All the cows have 4 legs and all the chickens have 2 legs.

lora16 [44]

Answer:

21 chickens

Step-by-step explanation:

10×4=40

82-40=42

42÷2=21

3 0

3 years ago

Please help answer please

amid [387]

Answer:

x = 19

Step-by-step explanation:

x-42+x-32+180-x+17 = 180

3x +123 = 180

3x = 57

x=19

8 0

3 years ago

Read 2 more answers

The measure of the largest angle of a triangle is 80º more than the measure of the smallest angle, and the measure of the remain

Korolek [52]

L = S + 80
180 - L - S = S + 10 simplify: L = 180 - 2S - 10
L = 170 - 2S = S + 80
3S = 90
S = 30 L = 110 3rd angle = 40

4 0

3 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