Ask question

Ask question

All categories

Katyanochek1 [597]

4 years ago

6

Graph the inequality

1 answer:

julsineya [31]4 years ago

3 0

-6 or could be -8..........hope i helped

You might be interested in

Helpppp me plz, algebra 1.

stiks02 [169]

Answer:

estimated product = 80, exact product = 79.79

6 0

3 years ago

Read 2 more answers

The value of x in the expression<br> 2x/3 - 7 = 5 is

Andrej [43]

Answer:

The answer is X = 18

Step-by-step explanation:

2x/3 - 7 = 5

Multiply both sides if the equation by 3 : 2x - 21 = 15

Move the constant to the right hand side and change its sign : 2x = 15 + 21

Calculate 2x = 15 + 21

2x = 36

Divide both sides of the eqaution by 2 : 2x = 36

x = 18

Answer X = 18

3 0

3 years ago

Explain how to write a whole number as a fraction with a denominator of 1

Troyanec [42]

Answer:

All you need to do is write the whole number and then a slash with the denominator of 1. For example: you would write 5 as 5/1 in fraction form

Step-by-step explanation:

6 0

3 years ago

Which of the following is equivalent 2x+4y

Fudgin [204]

Answer: None of them cuz

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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