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3 years ago

If both pairs of opposite angles of a quadrilateral are supplemental then the quadrilateral is a parallelogram

Mathematics

1 answer:

Nutka1998 [239]3 years ago

8 0

Answer:

THEOREM: If a quadrilateral has consecutive angles which are supplementary, then it is a parallelogram. BRAINLIEST LOL :)

Step-by-step explanation:

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Explain how you can use doubles when multiplying with 4 to find 4x8

hoa [83]

When multiplying by 4, you can simply multiply it by 2 twice (prime factorization).
In this example-
4 times 8 = 32
2 times 2 times 8 = 32

Hope this helped!
By the way, I'm getting closer to leveling up, so if you liked my answer, you could reward me with brainliest answer. That'd be great!

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3 years ago

You want to estimate the number of students in a junior high school who ride the school bus. which sample is unbiased?

Vika [28.1K]

Answer:

Option B - 90 students at random during lunch

Step-by-step

An unbiased sample is defined as:

A sample is an unbiased sample if every individual or the element in the population has an equal chance of being selected.

Based on this, we can eliminate:

Option A) - It prevents non-eighth-graders of the school from being represented.

Option C) - Students in a hallway might take a particular class together/may be of the same grade. This gives them an unfair advantage.

Option D) - All students who are not in chorus are not represented in this set.

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3 years ago

Can anyone help me with primary 5 hw ?

Illusion [34]

B) 28×50, then take the answer - 1100

5 0

3 years ago

EXPLAIN why we placed the value of x= 4/3( the minimum value) into the equ of gradient(dy/dx) [in the answer, marking scheme att

aliina [53]

$y=x(x-2)^2$
$\implies y'=(x-2)^2+2x(x-2)=3x^2-8x+4=(3x-2)(x-2)=0$
$\implies x=\dfrac23,x=2$

are the critical points, and judging by the picture alone, you must have $b=\dfrac23$ and $a=2$ . (You might want to verify with the derivative test in case that's expected.)

Then the shaded region has area

$\displaystyle\int_0^2x(x-2)^2\,\mathrm dx=\dfrac43$

I'll leave the details to you.

Now, for part (iv), you're asked to find the minimum of $\dfrac{\mathrm dy}{\mathrm dx}=y'$ , which entails first finding the second derivative:

$y'=3x^2-8x+4$
$\implies y''=6x-8$

setting equal to 0 and finding the critical point:

$6x-8=0\implies x=\dfrac86=\dfrac43$

This is to say the minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ *occurs when $x=\dfrac43$ *, but this is not necessarily the same as saying that $\dfrac43$ is the actual minimum value.

The minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ is obtained by evaluating the derivative at this critical point:

$m=\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=4/3}=3\left(\dfrac43\right)^2-8\left(\dfrac43\right)+4=-\dfrac43$

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4 years ago

Pleeeeeeeeese help quik

Ilya [14]

Answer:

Step-by-step explanation:

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3 years ago

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