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2 years ago

Find the slope (-2,1)and (1,2.5)

Mathematics

1 answer:

musickatia [10]2 years ago

8 0

Answer:

Step-by-step explanation:

slope = rise/run= 1.5/ 3= 0.5

hope that helps if it does please mark as brainliest.

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Which place value should you use to compare the numbers?<br><br> 645,853<br> 646,34

Ludmilka [50]

I’m not sure I understand the question

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2 years ago

What does x equal? plz help

erma4kov [3.2K]

Answer:

Step-by-step explanation:

∠E = ∠B

6x - 39 = 3x

6x - 3x = 39

3x = 39

x = 39/3

x = 13

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3 years ago

The student council needs to raise at least $950 to pay for their trip to Washington, DC. They have raised $589.55

podryga [215]

Answer:

181 bagels

Step-by-step explanation:

Well how much money did they have to raise more to get to $950? It is $360.45. So now we take $360.45 and divide it by 2 and we get 180.225. But if we round 180.225 they would have to sell at least 181 bagels to raise $950 dollars.

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2 years ago

Answer in 5 min I GIVE U BRAINLIEST

telo118 [61]

This is the formula of finding the hypotenuse: a² + b² = c².

So the legs would be a and b.

4² + 4² = c²

c² = 16 + 16

c² = 32

c = √32 (Option C)

7 0

2 years ago

Read 2 more answers

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

3 years ago