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2 years ago

Which pairs of quadrilaterals can be shown to be congruent using

Mathematics

1 answer:

Zielflug [23.3K]2 years ago

3 0

Rigid transformation do not alter the sides and angle measurements of a shape.

The true statements are:

quadrilateral 1 and quadrilateral 2 - Not congruent
quadrilateral 1 and quadrilateral 3 - Not congruent
quadrilateral 1 and quadrilateral 4 - Not congruent
quadrilateral 2 and quadrilateral 3 - Congruent

From the graph, we have the following highlights

Quadrilaterals 2, 3 and 4 are congruent
Quadrilateral 1 does not have equal measure with any of the other quadrilaterals.

Hence, the first three statements are not congruent, while the last statement is congruent.

Read more about congruent shapes at:

brainly.com/question/24430586

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A town was concerned about the growing number of harbor seals in their town's bay. They tagged 15 seals and sent them back into

Zinaida [17]

Answer:

im thinking 200 seals

Step-by-step explanation:

i don't have time sorry but you are welcome

3 0

3 years ago

Read 2 more answers

Your answer What is the greatest common factor of 12 and 24? Your answer 6

anastassius [24]

Answer:

yes it is 6

Step-by-step explanation:

Both are even numbers and caan be divided by 6 win 24 is divided by two I equals 12.

8 0

3 years ago

A container initially containing10 L of water in which there is 20 g of salt dissolved. A solution containing 4 g/L of salt is p

Kay [80]

Answer:

$A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196$

Step-by-step explanation:

For this case the solution flows at a rate of 2L/min and leaves at 1L/min. So then we can conclude the volume is given by $V= 10 +t$

Since the initial volume is 10 L and the volume increase at a rate of 1L/min.

For this case we can define A as the concentration for the salt in the container. And for this case we can set up the following differential equation:

$\frac{dA}{dt}= 4 \frac{gr}{L} *2 \frac{L}{min} - \frac{A}{10+t}$

Because at the begin we have a concentration of 8 gr/L and would be decreasing at a rate of $\frac{A}{10+t}$

So then we can reorder the differential equation like this:

$\frac{dA}{dt} +\frac{A}{10+t} =8$

We find the solution using the integration factor:

$\mu = -\int \frac{1}{10+t} dt = -ln(10+t)$

And then the solution would be given by:

$A = e^{-ln (10+t)} (\int e^{\int \frac{1}{10+t} dt})$

And if we simplify this we got:

$A= \frac{1}{10+t} (c + \int (10 +t) 8 dt)$

And after do the integral we got:

$A= \frac{c}{10+t} +4 (10 +t)$

And using the initial condition t=0 A= 20 we have this:

$20 = \frac{c}{10} +40$

$c= -200$

So then we have this function for the solution of A:

$A= \frac{-200}{10+t} +4 (10 +t)$

And now replacinf t= 40 we got:

$A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196$

8 0

3 years ago

Need points ?Answer question simple.

Tresset [83]

Answer:

45 degrees

Step-by-step explanation:

A straight line is an angle with a measure of 180°, but becasue it looks like we have two angles that create the 180° angle, therefore, it is a supplementary angle.

Knowing this

2X = 90° (right angle)

so (2X) + (2X) = 180° (supplementary angle)

Therefore, to find X you can do

2X = 90

divide by 2 on both sides (to isolate the variable we are trying to find which is X)

and you get....

X= 45 becasue 2 times 45 equals 90

answer: X=45°

7 0

3 years ago

Convert the following sum into a product. cosx + cos3x + cos5x + cos7x

kvv77 [185]

Answer:

$4\cos x\cos (4x)\cos (2x)$

Step-by-step explanation:

Given: $cosx + cos3x + cos5x + cos7x$

To convert: the given sum into product

Solution:

Use formula: $\cos A+\cos B=2\cos \left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )$

$cosx + cos3x + cos5x + cos7x=2\cos \left ( \frac{x+3x}{2} \right )\cos \left ( \frac{x-3x}{2} \right )+2\cos \left ( \frac{5x+7x}{2} \right )\cos \left ( \frac{5x-7x}{2} \right )\\=2\cos (2x)\cos (-x)+2\cos (6x)\cos (-x)\\=2\cos (2x)\cos (x)+2\cos (6x)\cos (x)\\=2\cos x\left [ \cos (2x)+\cos (6x) \right ]$

$cosx + cos3x + cos5x + cos7x=2\cos \left ( \frac{x+3x}{2} \right )\cos \left ( \frac{x-3x}{2} \right )+2\cos \left ( \frac{5x+7x}{2} \right )\cos \left ( \frac{5x-7x}{2} \right )\\=2\cos x\left [ \cos (2x)+\cos (6x) \right ]\\=2\cos x\left [2 \cos \left ( \frac{2x+6x}{2} \right )\cos \left ( \frac{2x-6x}{2} \right ) \right ]\\=2\cos x\left [ 2\cos (4x) \cos (-2x) \right ]\\=4\cos x\cos (4x)\cos (2x)$

4 0

3 years ago