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ser-zykov [4K]
2 years ago
11

PLEASE HELP ME FIND THE CORDINATES FOR THESE GRAPHSSS PLEASE ILL GIVE YOU BRAINLIEST AND 25 POINTS

Mathematics
1 answer:
tia_tia [17]2 years ago
7 0
1. (-1,-1) (2,-2)
2. (-3,0) (2,0)
3. (-4,-3) (-1,2)
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6 ft by 3 ft

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Tina’s bedroom is in the shape of a rectangular prism 15 feet long, 12 feet wide, and 10 feet high. It has no windows. Tina want
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720ft²

Step-by-step explanation:

Wall 1&2 = lh = 15*10=150

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7 0
3 years ago
Help plz!! i am having a hard time
beks73 [17]
If x represents the number of flower baskets, then, just as you showed, the answer cant be either a or b because store A doesnt charge $15 per flower basket but rather $15 for delivery. To see whether it is c or d, simply use the same number to represent the number of flower baskets in both inequalities. The easiest would be 1 because then youre multiplying - in the case of store A - 19.25 (how much each flower basket costs) times 1 (the total number of flower baskets) and then adding 15 (the cost of delivery) which is a total of 34.25. For store B it would be 17.5 times 1 plus 30 which equals 47.5. Therefore, the inequality for store A is less than that of store B, so D would be the answer. (i hope that sort of makes sense sjdjdjjs)
8 0
3 years ago
1.- What is 'a' for this hyperbola?
SpyIntel [72]

Answer:

The standard form of a hyperbola with vertices and foci on the x-axis:

\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1

where:

  • center:  (h, k)
  • vertices:  (h+a, k) and (h-a,k)
  • Foci:  (h+c, k) and (h-c, k) where the value of c is c² = a² + b²
  • Slopes of asymptotes: \pm\left(\dfrac{b}{a}\right)

<h3><u>Part 1</u></h3>

The center of the given hyperbola is (0, 0), therefore:

\implies \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1

Therefore (\pm a,0) are the vertices.  From inspection of the graph, a=2.

<h3><u>Part 2</u></h3>

Choose two points on the asymptote with the positive slope:

(0, 0) and (4, 6)

Use the slope formula to find the slope:

\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{6-0}{4-0}=\dfrac{3}{2}

<h3><u>Part 3</u></h3>

Use the <u>slopes of asymptotes</u> formula, compare with the slope found in part 2:

\implies \dfrac{b}{a}=\dfrac{3}{2}

Therefore, b=3

<h3><u>Part 4</u></h3>

Substitute the found values of a and b into the equation from part 1:

\implies \dfrac{x^2}{2^2}-\dfrac{y^2}{3^2}=1

\implies \dfrac{x^2}{4}-\dfrac{y^2}{9}=1

4 0
1 year ago
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