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4 years ago

One way a parallelograhm and a rhombos are diffrent

1 answer:

7 0

A parallelogram does not need to have all its sides equal. It just needs two pairs of equal opposite sides. A rhombus on the other hand, has to have all its four sides equal. A rhombus is actually a special form a parallelogram. So, a rhombus is always a parallelogram but a parallelogram is not always a rhombus.

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I’m so confused pls help...

Mazyrski [523]

Answer:

D) 24

<u>Steps:</u>

(2/10)=3/(x-9)

》1/5=3/(x-9)

》(x-9)=15

》x = 15+9

》x = 24

7 0

3 years ago

x is a point in the interior of rectangle PTRS. If XP=a, XT=c and XR=b, determine XS. a<b<c. Answer should be in terms of

White raven [17]

So you cannot really get the exact measure because they didn't give you any measures, but we can infer that if c>b and a then
maybe SX (d) is more than b and a so

a<b<c or d not sure which
sorry if I didn't help

3 0

3 years ago

Dave loves a bargain and buys a feather boa which

Ivanshal [37]

Answer:

£9.50

Step-by-step explanation:

if the price was reduced 70%, then Dave is paying 30% of the original.

0.3 of £ = 2.85

£ = 2.85 / 0.3

£ = 9.5

£9.50

3 0

4 years ago

Read 2 more answers

Wastewater is filling barrels at the rate of 11 quarts per hour. The recycling facility picks up 120 full barrels on each trip,

dalvyx [7]

(1 h)/(11 qt) * (12 qt)/(1 barrel) * (120 barrels/pickup) * (1 day)/(24 h)
.. = 12*120/(11*24) day/pickup
.. = 5 5/11 days/pickup

120 barrels will be filled every 5 5/11 days. Some trips will be 5 days apart, some trips will be 6 days apart.

8 0

3 years ago

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

3 0

3 years ago