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3 years ago

explain how you could use the construction tool or a compass and straightedge to create a line segment twice as long as AB

Mathematics

1 answer:

Hatshy [7]3 years ago

3 0

Is there a graph to this

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# 14 Q. Write a congruence

MrRissso [65]

The first is the correct
Δ ZYB ≈ Δ OWR by SSS
because:
                ZB = OR = 5
                ZY = OW = 3
                YB = WR = 5.5

4 0

3 years ago

Read 2 more answers

Water tank holds 711 gallons but is leaking at a rate of 4 gallons per week. second water tank holds

Shalnov [3]

After 79 weeks the tanks will have the same amount of water.
948-7*79=395
711-4*79=395

7 0

3 years ago

Can Mario organize his marble in equal piles

Rama09 [41]

No he cannot bc 61 is an odd number

4 0

3 years ago

Read 2 more answers

Henry spent 1 hour 30 minutes less than Joey reading last week. Joey spent 50 minutes less than Pete. Pete spent 3 hours reading

Nookie1986 [14]

Answer: Henry read for a total of 40 minutes

Step-by-step explanation: 3:00 - 0:50 is 2:10 then 2:10 - 1:30= 40 minutes

3 0

3 years ago

Read 2 more answers

On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes

belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

$r = \frac{\Delta T_{X}}{\Delta T_{Y}}$

$r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}$

$r = 11\,\frac{^{\circ}X}{^{\circ}Y}$

The difference between current temperature in Y linear scale with respect to freezing point is:

$\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)$

$\Delta T_{Y} = 115\,^{\circ}Y$

The change in X linear scale is:

$\Delta T_{X} = r\cdot \Delta T_{Y}$

$\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)$

$\Delta T_{X} = 1265\,^{\circ}X$

Lastly, the current temperature on the X scale is:

$T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X$

$T_{X} = 1150\,^{\circ}X$

The current temperature on the X scale is 1150 °X.

5 0

3 years ago