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2 years ago

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Determine if the statement is always, sometimes, or never true.The opposite angles of a parallelogram are congruent.alwayssometi

mesnever

2 answers:

ehidna [41]2 years ago

3 0

That's true. Opposite sides of a parallelogram are parallel, which implies that the opposite angles are congruent.

n200080 [17]2 years ago

3 0

Answer: Always true.

Step-by-step explanation:

Given statement : The opposite angles of a parallelogram are congruent.

A parallelogram is a kind of quadrilateral having four sides in which opposite sides and angles are congruent.

Therefore, the given statement " The opposite angles of a parallelogram are congruent." is always true by the definition of parallelogram.

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NEED Help!<br> 111=14g<br> and also 13.65=h + 4.88

crimeas [40]

Answer:

In this question if g and h is constant then;

111 = 14g

14g = 111

g = 111/14

g = 27.7 ~ 28

And also,

13.65 = h + 4.88

h + 4.88 = 13.65

h = 13.65 - 4.88

h = 8.77

Hope it helps!

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2 years ago

a construction company charges $15 per hour for debris removal , plus a one -time fee for the use of the trash dumpster.the tota

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Let h be the numbers of hours, f be the one-time fee and c the cost charged. The equation is

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Since you pay $15 per four, plus the fee. We can solve this equation for the fee:

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If we plug c=195 and h=9, we have

$f=195-15\cdot 9=195-135=60$

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2 years ago

Read 2 more answers

A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated tha

8_murik_8 [283]

To model this situation, we are going to use the decay formula: $A=Pe^{rt}$
where
$A$ is the final pupolation
$P$ is the initial population
$e$ is the Euler's constant
$r$ is the decay rate
$t$ is the time in years

A. We know for our problem that the initial population is 1,250, so $P=1250$ ; we also know that after a year the population is 1000, so $A=1000$ and $t=1$ . Lets replace those values in our formula to find $r$ :
$A=Pe^{rt}$
$1000=1250e^{r}$
$e^{r}= \frac{1000}{1250}$
$e^{r}= \frac{4}{5}$
$ln(e^{r})=ln( \frac{4}{5} )$
$r=ln( \frac{4}{5} )$
$r=-02231$

Now that we have $r$ , we can write a function to model this scenario:
$A(t)=1250e^{-0.2231t}$ .

B. Here we are going to use a graphing utility to graph the function we derived in the previous point. Please check the attached image.

C.
- The function is decreasing
- The function doe snot have a x-intercept
- The function has a y-intercept at (0,1250)
- Since the function is decaying, it will have a maximum at t=0:
$A(0)=1250e^{(-0.2231)(0)$
$A_{0}=1250e^{0}$
$A_{0}=1250$
- Over the interval [0,10], the function will have a minimum at t=10:
$A(10)=1250e^{(-0.2231)(10)$
$A_{10}=134.28$

D. To find the rate of change of the function over the interval [0,10], we are going to use the formula: $m= \frac{A(0)-A(10)}{10-0}$
where
$m$ is the rate of change
$A(10)$ is the function evaluated at 10
$A(0)$ is the function evaluated at 0
We know from previous calculations that $A(10)=134.28$ and $A(0)=1250$ , so lets replace those values in our formula to find $m$ :
$m= \frac{134.28-1250}{10-0}$
$m= \frac{-1115.72}{10}$
$m=-111.572$
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.

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Arada [10]

Answer:

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Step-by-step explanation:

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