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gladu [14]
3 years ago
15

Use a unit rate to find the unknown value.

Mathematics
1 answer:
Tanzania [10]3 years ago
3 0
1. 40/8 = 45/9
2. 42/14 = 15/5
3. 14/2 = 56/8
4. 8/4 = 26/13
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If m (AB = 120°, then which of the following have measures of 60°?
victus00 [196]
<span>∠BCA = 120 /2 = 60

answer
</span><span>∠BCA</span>
7 0
2 years ago
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A, 5+x = 17
Vlad [161]

A. 5+x=17, x = 17-5=12

B. x = 4 x 2.5 = 10

C. x = 27+12 = 39

D. x = 16:4 = 4

E. x = 9 : 3/4 = 12

Answer A and E

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2 years ago
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Find the area of the figure. Round to the nearest tenth if necessary.
r-ruslan [8.4K]
Area =  area of rectangle - area of 2 hemispheres
         = 24 * 32 - pi * 12^2
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5 0
3 years ago
An engineer is designing a large steel pad to be installed on the deck of an aircraft carrier. Its total volume will be 18 yd^3.
Shtirlitz [24]

Answer:

The cost of the material for the full-sized pad is;

d) 222,750.00

Step-by-step explanation:

The parameters of the steel pad and the scale model are;

The volume of the deck = 18 yd³

The dimensions of the model of the same material = 6 feet by 4.5 feet and 4.5 inches thick

The weight of the sample, m₂ = 75 lbs

The cost of the steel, P = $55/lb

The dimensions of the sample in yards are;

6 feet = 2 yards, 4.5 feet = 1.5 yards, and 4 inches = 0.\overline 1 yards

The volume of the sample, V₂ = 2 yd. × 1.5 yd. × 0.\overline 1 yd. =  (1/3) yd.³

The density of the sample material, ρ = m₂/V₂

∴ The density of the sample material, ρ = 75 lbs/((1/3)yd.³) = 225 lbs/yd³

Therefore, the density of the material of the pad, ρ = 225 lbs/yd³

The mass of the steel  pad, m₁ = ρ × V₁

∴ The mass of the steel  pad, m₁ = 225 lbs/yd³ × 18 yd³ = 4,050 lbs

The cost of the material for the full-sized steel pad, C = m₁ × P

∴ C = 4,050 lbs × $55/lb = $222,750

6 0
3 years ago
GraceRosalia only should answer: No one else ....I beg others not to answer..<br>√625 - √25 + √100​
artcher [175]

\sqrt{625} =25

√25=5

√100=10

25-5+10=30

Good luck

ANSWERED:\\GraceRosalia

8 0
2 years ago
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