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4 years ago

Which postulate or theorem proves ∆MNQ ≅ ∆PNQ ?

Mathematics

2 answers:

Finger [1]4 years ago

6 0

ASA Congruence Postulate

AysviL [449]4 years ago

4 0

AAS Congruence Postulate

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A choir teacher is dividing 12 sopranos and 9 altos into singing groups. He wants each group to have the same combination of sop

almond37 [142]

Answer:

3 groups

Step-by-step explanation:

Sopranos = 12

Altos = 9

Find the highest common factor of 12 and 9

12 = 1, 2, 3, 4, 6, 12

9 = 1, 3, 9

The highest common factor of 12 and 9 is 3

Therefore, the choir teacher can only have 3 groups

12 sopranos / 3 groups

= 4 sopranos per group

9 Altos / 3 groups

= 3 altos per group

Each of the 3 groups Will have 4 sopranos and 3 altos each

3 0

3 years ago

1. the height of a water bottle in a bathtub, w, is a function of time, t. Let P represent this function. Height is measured in

zhannawk [14.2K]

3. After 10 minutes, the height of the water is 4 inches

Answer:

1) P(20) = 0

2) P(0) = 0

3) P(10) = 4

4) P(4) = 10

Step-by-step explanation:

We are told that the function P(t) denotes the height of the water in the bath tub.

At P(0) = 0, it means that after 0 minutes, there is no water in the bath tub.

1) If after 20 minutes, the bathtub is empty, then we can depict the function as; P(20) = 0

2) The bath tub starts out with no water. It means at 0 minutes, there was no water in the tub. Thus;

P(0) = 0

3) If after 10 minutes, the height of the water is to be 4 inches, then;

P(10) = 4

4) If the height is 10 inches after 4 minutes, then;

P(4) = 10

4 0

3 years ago

A spherical balloon currently has a radius of 19cm. If the radius is still growing at a rate of 5cm or minute, at what rate is a

miv72 [106K]

Answer:

22670.8 cm³/min

Step-by-step explanation:

Given:

Radius of the balloon at a certain time (r) = 19 cm

Rate of growth of radius is, $\frac{dr}{dt}=5\ cm/min$

The rate at which the air is pumped in the balloon can be calculated by finding the rate of increase in the volume of the balloon.

So, first we find the volume of the sphere in terms of 'r'. As the balloon is spherical in shape, the volume of the balloon is equal to the volume of a sphere. Therefore,

Volume of balloon is given as:

$V=\frac{4}{3}\pi r^3$

Now, rate of increase of volume is obtained by differentiating both sides of the equation with respect to time 't'.

Differentiating both sides with respect to time 't', we get:

$\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4\pi}{3}(3r^2)(\frac{dr}{dt})\\\\\frac{dV}{dt}=4\pi r^2(\frac{dr}{dt})$

Now, plug in 19 cm for 'r', 5 cm per minute for $\frac{dr}{dt}$ and solve for $\frac{dV}{dt}$ . This gives,

$\frac{dV}{dt}=4\pi (19 cm)^2(5\ cm/min)\\\\\frac{dV}{dt}=4\times 3.14\times 361\times 5\ cm^3/min\\\\\frac{dV}{dt}=22670.8\ cm^3/min$

Therefore, the rate at which the air is being pumped into the balloon is 22670.8 cm³/min.

4 0

4 years ago

Diego’s family car holds 14 gallons of fuel. Each day the car uses 0.6 gallons of fuel. A warning light comes on when the remain

Tomtit [17]

No the could not go 25 days without it coming on....if you multiply.6 gallons by 25 days you get 15 since the car only holds 14 gallons the car would be empty before the 25th day

8 0

3 years ago

Read 2 more answers

Solve for x. (x - 8)(x - 8) = 0

raketka [301]

The answer is x equals 8

3 0

3 years ago

Read 2 more answers