All categories

3 years ago

Hexagon DEFGHI is translated on the coordinate plane below to create hexagon D'E'F’G’H’I’

Mathematics

1 answer:

frez [133]3 years ago

7 0

Answer:

The answer is " $\bold{(x,y)\longrightarrow (x-8, y-7)}$ ".

Step-by-step explanation:

In the given question some data is missing. so, the correct answer to this can be explained in the following example:

In the given example, point D is also known as the coordinates in the (2, 5), and in the coordination of the D prime (D') and after translation its co-ordinates value is (-6, -2). Since the very first choice is only accurate as:

⇒2-8 = -6

⇒ 5-7 = -2

It can try certain things too and see why this alone is the correct translation.

You might be interested in

Hey! Giving away 50 free points! Just type "octopus" as an answer and I will give you 5 star, a heart, and brainly will give you

Arturiano [62]

Answer:

octopus

Step-by-step explanation:

thank you so muchhhhh

3 0

3 years ago

Read 2 more answers

17. Four identical bags of cement and a

pogonyaev

The mass of each bag of cement is 16.5 kg. Check photo for steps and details, hope this helped! Please mark brainliest if you think this was helpful c:

5 0

3 years ago

Could you plz help me on my hw

Korolek [52]

7. p=0.4
i dont know about 3
or 6

5 0

3 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

3 years ago

An arithmetic sequence with a thrid term of 8 and constant difference of 5

olga nikolaevna [1]

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=5\\
a_3=8\\
n=3
\end{cases}
\\\\\\
a_3=a_1+(3-1)5\implies 8=a_1+(2)5\implies 8=a_1+10
\\\\\\
\boxed{-2=a_1}
\\\\\\
a_n=-2+(n-1)5$

8 0

4 years ago