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SVEN [57.7K]
3 years ago
5

What are the coordinates of the point (-3,4) after a translation of 2 units down and 5 units left

Mathematics
1 answer:
marta [7]3 years ago
5 0

Answer:

(-8, 2)

Step-by-step explanation:

Start with (-3, 4).  Subtract 5 units from -3, obtaining -8, and subtract 2 units from 4, obtaining 2:  (-8, 2)

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A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 5 − x 2 . What are the dimensions
kherson [118]

Answer:

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola

y=5−x^2. What are the dimensions of such a rectangle with the greatest possible area?

Width =

Height =

Width =√10 and Height = \frac{10}{4}

Step-by-step explanation:

Let the coordinates of the vertices of the rectangle which lie on the given parabola y = 5 - x² ........ (1)

are (h,k) and (-h,k).

Hence, the area of the rectangle will be (h + h) × k

Therefore, A = h²k ..... (2).

Now, from equation (1) we can write k = 5 - h² ....... (3)

So, from equation (2), we can write

A =h^{2} [5-h^{2} ]=5h^{2} -h^{4}

For, A to be greatest ,

\frac{dA}{dh} =0 = 10h-4h^{3}

⇒ h[10-4h^{2} ]=0

⇒ h^{2} =\frac{10}{4} {Since, h≠ 0}

⇒ h = ±\frac{\sqrt{10} }{2}

Therefore, from equation (3), k = 5 - h²

⇒ k=5-\frac{10}{4} =\frac{10}{4}

Hence,

Width = 2h =√10 and

Height = k =\frac{10}{4}.

8 0
3 years ago
I’m not sure if i am correct with my answer. Can Someone help or confirm which answer it is or if i’m correct or not? :)
Alex
Hi! i think it’s the last one (270 degrees counterclockwise and 1 unit to the right)

if it was 180 degree, the shape would be in quadrant II, not quadrant III
8 0
3 years ago
How do you combine like terms in this expression? <br><br> 10xy-4(xy+2x²y)
klio [65]
The answer is - 8x^2y+6xy
7 0
3 years ago
We need to take care of these annoying bots once and for all
satela [25.4K]

Answer:

yes we should >:(

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Find the missing value. Show your work. please
Sophie [7]
#1
sin(51^o)= \frac{y}{12} \ \ \to \ \ y=12*sin(51^o)=12* 0.7771 \approx 9.33

#2
\theta=arccos( \frac{5^2+13^2-12^2}{2*5*13}) = arccos( \frac{25+169-144}{130})=arccos( \frac{50 }{130})= \\ \\ arccos(0.3846) \approx67.4^o

#3
tan(13^o)= \frac{x}{24} \ \ \to \ \ x=24*tan(13^o)=24* 0.2309 \approx 5.54

#4
sin(20^o)= \frac{10}{x} \ \ \to \ \ x= \frac{10}{sin(20^o)} = \frac{10}{0.342}  \approx 29.24
5 0
3 years ago
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