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1 year ago

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the distance between california and mexico border is 100 miles less than thrice the distance between california and washington.

if m is the distance between california and washington, which expression can be used to denote distance between california and mexico. a.100-2m b. 3m-100 c. 100m-3 d. 3-100m

1 answer:

harina [27]1 year ago

3 0

Answer: B. 3m-100

Step-by-step explanation:

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s2008m [1.1K]

Answer:

4

Step-by-step explanation:

2×4 is the only one that equals less than 9, 2× 4.5 is 9, 2×5.5 is 11, 2×6 is 12, and the question is trying to get below 9, so 4 is the only answer that can get below 9

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3 years ago

Please help me with this

Cerrena [4.2K]

Answer:

Yes;Each side of triangle PQR is the same length as the corresponding side of triangle STU

Step-by-step explanation:

You can observe the sides of both triangles to see if this property holds

Lets check the length of AB

A(0,3) and B(0,-1)

$AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\\\\\\AB=\sqrt{(0-0)^2+(-1-3)^2} \\\\\\\\AB=\sqrt{0^2+-4^2} \\\\\\AB=\sqrt{16} =4units$

Now check the length of the corresponding side DE

D(1,2) and E(1,-2)

$DE=\sqrt{(1-1)^2+(-2-2)^2} \\\\\\DE=\sqrt{0^2+-4^2} \\\\\\DE=\sqrt{16} =4units$

The side AB has the same length as side DE.This is also true for the remaining corresponding sides.

6 0

3 years ago

How can I figure out this math problem: Stanley ran a 5 kilometer race in 30 minutes. What was his speed in kilometers per hour?

stiv31 [10]

Since the question is asking for the time units to be in hours, you want to convert 30 minutes to an hour; 0.5

We’re able to use the basic motion formula for velocity to work out this question.

Velocity = distance / time
Velocity = 5 / 0.5
Velocity = 10km/h

Therefore, Stanley’s average speed was 10 kilometres per hour. :)

3 0

2 years ago

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.

7 0

3 years ago

HELP PLEASE!!!!!!!!!! ASAP

Lilit [14]

Answer:

j - 150

k - 62

l - 52

m - 96

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3 years ago