All categories

3 years ago

Find the vertex of the parabola.

Mathematics

1 answer:

7nadin3 [17]3 years ago

7 0

Answer:

C: (3,1)

Step-by-step explanation:

y = -4x² + 24x - 35

y = -4(x² - 6x) - 35

y = -4[x² - 2(x)(3) + 3² - 3²] - 35

y = -4(x - 3)² -4(-9) - 35

y = -4(x - 3)² + 36 - 35

y = -4(x - 3)² + 1

Vertex (3,1)

You might be interested in

The baseball team needs new equipment. Company A can provide 9 helmets, 6 bats, and 12 balls for $525. Company B can provide 10

Olin [163]

Answer:

Choice A.

Step-by-step explanation:

Let x = price of a helmet, y = price of a bat, z = price of a ball.

Company A can provide 9 helmets, 6 bats, and 12 balls for $525.

9x + 6y + 12z = 525

Company B can provide 10 helmets, 8 bats, and 10 balls for $600.

10x + 8y + 10z = 600

Company C can provide 8 helmets, 5 bats, and 15 balls for $500.

8x + 5y + 15z = 500

Answer: Choice A.

6 0

3 years ago

Read 2 more answers

Please answer this question right now

likoan [24]

Answer:x= 0.5,−3

Step-by-step explanation:

4 0

2 years ago

Jimmy wants to order pizza for the birthday party on Thursday each pizza has four slices if the eight kids at the party each eat

blagie [28]

Answer: 14 Pizzas

2 pizzas means each kid gets one

so 7 times 2 is 14 so I beleive the answer is 14

8 0

2 years ago

Ax + b2 = cx + d2<br> ????

ryzh [129]

Answer:

x=b^2+d^2/a-c

Step-by-step explanation:

5 0

2 years ago

You have $7000 with which to build a rectangular enclosure with fencing. The fencing material costs $30 per meter. You also want

Arlecino [84]

Answer:

1857.12 square meters

Step-by-step explanation:

Let L be the length of the rectangle and 'W' be the width

Perimeter = $2L + 2W + 2W$

The fencing material costs $30 per meter.

The material for the partitions costs $25 per meter

$cost=30(2L + 2W) +25( 2W)\\60L+60W+50W\\60L +110W$

$7000=60L+110W$

Solve for L

$7000=60L+110W\\7000-110W= 60L\\L=\frac{700}{6} -\frac{11}{6} W$

Area = length times width

$( \frac{700}{6} -\frac{11}{6} W)(W)\\A(W)=\frac{700}{6}W -\frac{11}{6} W^2$

Now take derivative and set it =0

$A(W)=\frac{700}{6}W -\frac{11}{6} W^2\\A'(W)=\frac{700}{6} -\frac{22}{6} W$

set the derivative =0 and solve for W

$0=\frac{700}{6} -\frac{22}{6} W\\\frac{700}{6}=\frac{22}{6} W\\W= 31.8$

So width = 31.8 that gives maximum area

$L=\frac{700}{6} -\frac{11}{6} (31.8)=58.4$

$Area = length \cdot width = 31.8 \cdot 58.4= 1857.12$ square meter

4 0

3 years ago