All categories

3 years ago

Find each measurement. Assume that each figure is not drawn to scale.

Mathematics

1 answer:

ivanzaharov [21]3 years ago

4 0

Answer:

1) 2 1/4 inches (2.25)

2) 5.3 millimeters

3) X=10 and YZ=60

Step-by-step explanation:

Number 1, you just add the measurements because n and q is the start and end of the line segment.

Number 2, you are given the whole measurement, and you know FG which is 9.7, so you subtract the 15 (whole measurement) by 9.7, which is 5.3.

Number 3, you know that XY and YZ equal XZ, so 8x+1 would equal 81. Do the math, you would get 8x=80 and then x=8. Now you know x, plug it in for 6x, so 6(10)=60.

You might be interested in

Jorge has some dimes and quarters. He has 10 more dimes than quarters and the collection of coins is worth $2.40. How many dimes

PIT_PIT [208]

Answer:

14 dimes and 4 quarters

Step-by-step explanation:

x= number of dimes

y=number of quarters

x-10=y

0.10x+0.25y=2.40

substitute for y

0.10x+0.25(x-10)=2.40

0.35x-2.50=2.40

0.35x=4.90

x=14

y=4

5 0

3 years ago

Which of the following is a solution to the equation x + 16 = -14?

IRINA_888 [86]

Answer:

x = -30

Step-by-step explanation:

Step 1: Subtract 16 from both sides

x + 16 = -14

x + 16 - 16 = -14 - 16

x = -30

Answer: x = -30

7 0

3 years ago

Read 2 more answers

Decompose and use the distributive property to find the product of 5 × 7.3.

swat32

Answer:

36.5

Step-by-step explanation:

5(1+6.3)

6 0

3 years ago

two trains leave the station at the same time one heading west and the other East the westbound train travels at 55 miles per ho

Diano4ka-milaya [45]

Recall your d = rt, distance = rate * time

so... one train goes west and the goes the opposite way... alrite... so... notice, by the time 338 miles have been covered by both, it will have been "t" hours, and whatever "t" is, is the same amount of time the westbound train has been running as well as the eastbound train has been running.

now, let's say, since by "t" hours they've covered 338 altogether, so, if the westbound train has covered say "d" miles, then the eastbound train would have covered the slack from 338 and d, that is, "338 - d".

$\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
Westbound&d&55&t\\
Eastbound&338-d&75&t
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=55t\\
338-d=75t\\
----------\\
338-\boxed{55t}=75t
\end{cases}
\\\\\\
338=75t+55t\implies 338=130t\implies \cfrac{338}{130}=t\implies \stackrel{hours}{2.6}=t$

so, 2hours and 36 minutes.

8 0

3 years ago

A cable television compan is laying cable in an area with underground utilities. Two subdivisions are located on opposite sides

Schach [20]

Answer:

Step-by-step explanation:

Given:

Point Q = Q is on the north bank 1200 m east of P $40/m to lay cable underground and $80/m way to lay the cable.

Lets denote T be a point on north bank, therefore, point T is t m east and 100m north of P.

Note that 0 < t < 1200

Distance from P to T using Pythagoras theorem:

PT^2 = PQ^2 + QT^2

PT = √(t²+100²) = √(t²+10,000)

Distance from T to Q:

TQ = 1200 - t

From above, Cable from P to T is underwater, and costs $80/m to lay.

Cable from T to Q is underground, and costs $40/m to lay.

Total cost of the cable:

C(t) = 80√(t²+10,000) + 40(1200 - t)

Cost is at its minimum: when dC/dt = C'(t) = 0 and d^2C/dt^2 = C''(t) > 0

dC/dt = C'(t)

= 80 * 1/2 (t² + 10,000)^(-1/2) × 2t + 40 × (-1)

dC/dt = C'(t)

= 80t/√(t² + 10,000) - 40 = 0

80t/√(t² + 10,000) = 40

80t = 40√(t² + 10,000)

2t = √(t² + 10,000)

square both sides:

4t² = t² + 10,000

3t² = 10,000

t² = 10,000/3

t = 100/√3

≈ 57.735

d^2C/dt^2 =

C''(t) = [80√(t² + 10,000) - 80t × t/√(t²+10,000)] /√(t²+10,000)

C''(t) = [(80(t²+10,000) - 80t²) / √(t²+10,000)] /√(t²+10,000)

C''(t) = 800,000 / (t² + 10,000)^(3/2)

C''(t) > 0 for all t

Therefore C(t) is minimized at t = 100/√3

C(100/√3) = 80√(10,000/3 + 10,000) + 40 × (1200 - 100/√3)

= 80 √(40,000/3) + 48,000 - 4000/√3

= 16000/√3 + 48,000 - 4000/√3

= 48,000 + 12,000/√3

= 48,000 + 4,000√3

= 4000 (12 + √3)

≈ 54,928.20

Minimum cost is $54,928.20

This occurs when cable is laid underwater from point P to point T(57.735 m east and 100m north of P) and then underground from point T to point Q (1200 - 57.735)

= 1142.265 m

5 0

3 years ago