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3 years ago

6 pints of glaze it takes 7/8 of a long for a bowl now many bowls can you glaze

Mathematics

1 answer:

lubasha [3.4K]3 years ago

3 0

Answer:

ok so umm i dont get what ur looking for?

Step-by-step explanation:

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You earn 50,000 per year,and paid 10 percent in taxes this year. the government increased the tax rate to 20 percent foe the nex

bulgar [2K]

Answer:

Step-by-step explanation:

he paid 5000 in taxes last year multiply by 2 10000

D is the answer sorry i dont have a proper explanation

4 0

3 years ago

The mean weight 20 students in a room is 65 kg one student one student who weight is 88kg went into the room what is the new mea

umka2103 [35]

The new mean is 66.09

Explanation:

Mean weight of 20 students = 65kg

One student with 88kg entered the room.

New mean = ?

$\frac{x_1+x_2+....+x_2_0}{20} = 65$

$\frac{x_1+x_2+....+x_2_0 + 88}{21} = ?$

The above equation can further be written as:

$x_1 + x_2 +....+x_2_0 = 65 X 20\\\\x_1 + x_2 +....+x_2_0 = 1300$

New mean = $\frac{1300 + 88}{21}$

= 66.09

Therefore, the new mean is 66.09

6 0

3 years ago

A home is to be built on a 56 foot 9 inch wide lot. The house is 5 feet 5 inches from the side of the lot and is 34 feet 10 inch

daser333 [38]

Total lot - house -other side
=(56 ft 9 in) - (34 ft 10 in) - (5 ft 5 in)
=(55 ft 21 in) - (34 ft 10 in) - (5 ft 5 in)
=16 ft 6 in
Convert to inches: 16 ft x 12 in/ft)+ 6 in= Total in inches192 in + 6 in= Total in inches
198 in= Total in inches
Check: One side + house + other side = Lot width
(5 ft 5 in)+(34 ft 10 in)+(16 ft 6 in)=Lot width
56 ft 9 in=Lot width, which equals the given.

8 0

3 years ago

Prove that the triangle EDF is isosceles. Give reasons for your answer.

Gekata [30.6K]

Answer:

$\triangle EDF$ is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are given the following things:

$\angle EDF = y$

$\text{External }\angle DFG = 90 +\dfrac{y}{2}$

Let us try to find out $\angle E$ and $\angle DFE$ . After that we will compare them.

Finding  $\angle DFE$ :

Side EG is a straight line so $\angle GFE = 180$

$\angle GFE$ is sum of internal $\angle DFE$ and external $\angle DFG$

$\angle GFE = 180 = \angle DFE + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)$