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3 years ago

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Phoebe’s Bait and tackle sold 128,107 live worms in April and 102,278 live worms in may to the nearest thousand how many worms d

Id phoebe’s Bait and tackle sellin April and may together

1 answer:

amm18123 years ago

5 0

Phoebe’s Bait and tackle sold 128,107 live worms in April and 102,278 live worms in may to the nearest thousand how many worms did phoebe’s Bait and tackle sell in April and may together

Answer: Number of live worms sold in April by Phoebe's Bait and tackle $=128,107$

Number of live worms sold in May by Phoebe's Bait and tackle $=102,278$

Therefore, the number of live worms sold in April and May by Phoebe's Bait and tackle is:

$128,107 + 102,278=230,385$

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You are doing annual inventory at the hardware store where you work. A box filled with brass connectors weighs 5 pounds 4 ounces

Georgia [21]

Answer:

304 brass connectors.

Step-by-step explanation:

We have been given that a box filled with brass connectors weighs 5 pounds 4 ounces. The box weighs 8 ounces when empty, and each connector weighs 0.25 ounces.

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6 0

4 years ago

At the dog park on Monday, the ratio of small dogs to all dogs was three to seven. A total of 42 dogs were at the park. How many

faust18 [17]

Answer:

18 small dogs

Step-by-step explanation:

Given the ratio of small dogs to all dogs was 3:7

Total dogs that are there = 42 dogs

Number of small dogs there = ratio of small/ratio of all dogs * Total dogs

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3 years ago

HELP ASAP WILL MARK BRAINLIEST AREA OF FIGURES

Lerok [7]

Answer:

1. 170.083 in³

2. 126π in³

3. 92.106 m³

4. 2412.74 in³

5. 612π m³ and 1922 m³

Step-by-step explanation:

1.

Cylinder:

$V = \pi r^{2}h$ *Plug in numbers*

$(3.14)(2.5)^{2}(7)$ *Square 2.5*

$(3.14)(6.25)(7)$ *Solve*

≈ $137.375in^{3}$

Sphere:

$V = \frac{4}{3}\pi r^{3}$ *Plug in numbers*

$\frac{4}{3} (3.14)(2.5)^{3}$ *Cube 2.5*

$\frac{\frac{4}{3}(3.14)(15.625)}{2}$ *Divide by 2 and Solve*

≈ $32.7083 in^{3}$

Add both volumes

$137.375 + 32.7083$ ≈ $170.083in^{3}$

2.

Cylinder:

$V = \pi r^{2}h$ *Plug in numbers*

$\pi (3)^{2}(10)$ *Square 3*

$\pi (9)(10)$ *Multiply*

$90\pi$

Sphere:

$V = \frac{4}{3}$ π $r^{3}$ *Plug in numbers*

$\frac{4}{3}\pi (3)^{3}$ *Cube 3*

$\frac{4}{3} \pi (27)$ *Multiply*

$36\pi$

Add both Volumes to get total

$90\pi + 36\pi = 126in^{3}$

3.

Sphere:

$V = \frac{4}{3}\pi r^{3}$ *Plug in numbers*

$\frac{4}{3} (3.14)(3)^{3}$ *Cube 3*

$\frac{4}{3} (3.14)(27)$ *Multiply*

$113.04m^{3}$

Cone:

$V = \frac{\pi r^{2}h}{3}$ *Plug in numbers*

$\frac{(3.14)(2)^{2}(5)}{3}$ *Square 2*

$\frac{(3.14)(4)(5)}{3}$ *Solve*

$20.93m^{3}$

Subtract the volumes to get the volume of the blue area

$113.04 - 20.93 = 92.106m^{3}$

4.

Sphere:

$V = \frac{4}{3} \pi r^{3}$ *Plug in numbers*

$\frac{4}{3}\pi (8)^{3}$ *Cube 8*

$\\\frac{4}{3}\pi (512)$ *Multiply*

$\\\\\pi (682.6)$ *Solve*

$2133.66in^{3}$ *Divide by 2 since it's a hemisphere*

Cone:

$V = \frac{\pi r^{2}h}{3}$ *Plug in numbers*

$\frac{\pi (8)^{2}(20)}{3}$ *Square 8*

$\frac{\pi (64)(20)}{3}$ *Multiply and Divide*

$1340.41 in^{3}$

Add both volumes

$1072.33 + 1340.41 = 2412.74in^{3}$

5.

Cylinder:

$V = \pi r^{2}h$ *Plug in numbers*

$\pi (6)^{2}(16)$ *Square 6*

$\pi (36)(16)$ *Multiply*

$576\pi$

Cone:

$V = \frac{\pi r^{2}h}{3}$ *Plug in numbers*

$\frac{\pi (6)^{2}3}{3}$ *Square 6*

$36\pi$

Add both volumes

$576\pi + 36\pi = 612\pi m^{3}$

Alternative: *Multiply π*

$1922m^{3}$

4 0

3 years ago

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