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Maksim231197 [3]

1 year ago

13

A quadrilateral has vertices A(4, 5), B(2, 4), Q4, 3), and D(6, 4). Which statement about the quadrilateral is true?

1 answer:

Igoryamba1 year ago

7 0

The true statement about the quadrilateral is ABCD is a rhombus with non-perpendicular adjacent sides.

<h3>How to determine the true statement?</h3>

The coordinates are given as:

A(4, 5), B(2, 4), Q4, 3), and D(6, 4).

Next, we plot the graph of the quadrilateral ABCD

See attachment

From the attached graph, we have the following highlights:

Opposite sides are equal
Opposite sides are parallel
Adjacent sides are not perpendicular

This means that the quadrilateral is ABCD is a rhombus

Read more about quadrilateral at

brainly.com/question/4812739

#SPJ1

<u>Complete question</u>

A quadrilateral has vertices A(4, 5), B(2, 4), C(4, 3), and D(6, 4). Which statement about the quadrilateral is true?

1) ABCD is a parallelogram with noncongruent adjacent sides.

2) ABCD is a trapezoid with only one pair of parallel sides.

3) ABCD is a rectangle with noncongruent adjacent sides.

4) ABCD is a square.

5) ABCD is a rhombus with non-perpendicular adjacent sides.

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6 0

3 years ago

A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, wri

GuDViN [60]

Answer:

1. $\frac{dy}{dt}=ky$

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

$\frac{dy}{dt}$ =Number of bacteria per unit time

$\frac{dy}{dt}\proportional y$

$\frac{dy}{dt}=ky$

Where k=Proportionality constant

2. $\frac{dy}{y}=kdt$ ,y'(0)=100

Integrating on both sides then, we get

$lny=kt+C$

We have y(0)=200

Substitute the values then , we get

$ln 200=k(0)+C$

$C=ln 200$

Substitute the value of C then we get

$ln y=kt+ln 200$

$ln y-ln200=kt$

$ln\frac{y}{200}=kt$

$\frac{y}{200}=e^{kt}$

$y=200e^{kt}$

Differentiate w.r.t

$y'=200ke^{kt}$

Substitute the given condition then, we get

$100=200ke^{0}=200 \;because \;e^0=1$

$k=\frac{100}{200}=\frac{1}{2}$

$y=200e^{\frac{t}{2}}$

Substitute t=2

Then, we get $y=200e^{\frac{2}{2}}=200e$

$y=200(2.718)=543.6=543.6$

e=2.718

Hence, the number of bacteria after 2 hours=543.6

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lukranit [14]

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3 years ago

Let the function f be continuous and differentiable for all x. Suppose you are given that f(−1)=−3, and that f'(x) for all value

Monica [59]

Answer:

Step-by-step explanation:

By MVT,

$f'(c) = \frac{f(5)-f(-1)}{5-1}$

$4f'(c)= f(5) + 3\rightarrow f(5) = 4f'(c) - 3$

Moreover, $f'(x)\leq 4~~~~~~ \forall x,$

$f(5) \leq 16 - 3 =13$

Hence the largest possible value of f(5) is 13

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3 years ago

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3 years ago