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3 years ago

The PTA 125 donuts for $1.25 each. How much did they spend on Donuts?

Mathematics

2 answers:

AlladinOne [14]3 years ago

7 0

Multiply 1.25 by the 125 dounuts.

125
x 1.25

the answer is $156.25

Dominik [7]3 years ago

6 0

They spent $156.25 on donuts. Multiply 125*1.25.

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Please help solve -2x - 4 - 7 = - 3x - 15

valentina_108 [34]

Answer:

x = -4

Step-by-step explanation:

-2x-11=-3x-15

x - 11 = -15

x = -4

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3 years ago

AB and DE are chords that intersect at point F inside circle C as shown. If the measure of arc EA= 50 degrees and the measure of

defon

Answer:

Part 1) $m\angle AFE=55^o$

Part 2) $m\angle EFB=125^o$

Step-by-step explanation:

Part 1) what is the measure of angle AFE

we know that

The measure of the interior angle is the semisum of the arches that comprise it and its opposite.

Note: In this problem the correct measure of arc EA is 40 degrees (see the picture)

$m\angle AFE=\frac{1}{2}(arc\ EA+arc\ DB)$

substitute the given values

$m\angle AFE=\frac{1}{2}(40^o+70^o)=55^o$

Part 2) what is the measure of angle EFB?

we know that

$m\angle AFE+m\angle EFB=180^o$ ---> by supplementary angles (form a linear pair)

substitute the given value

$55^o+m\angle EFB=180^o$

$m\angle EFB=180^o-55^o=125^o$

8 0

3 years ago

Solve 3(x - 2) < 18. Please help me

lakkis [162]

Answer:

x > 8

Step-by-step explanation:

I hope this helps you out!

4 0

2 years ago

Read 2 more answers

Debra walked her dog to the south. They walked 50 meters in 150 seconds.

abruzzese [7]

Answer:

Step-by-step explanation:

Because... 150 divided by 50 = 3

4 0

2 years ago

How many different ways are there to choose a subset of the set {1,2,3,4,5,6} so that the product of the members of the subset i

zvonat [6]

You have to pick at least one even factor from the set to make an even product.

There are 3 even numbers to choose from, and we can pick up to 3 additional odd numbers.

For example, if we pick out 1 even number and 2 odd numbers, this can be done in

$\dbinom 31 \dbinom 32 = 3\cdot3 = 9$

ways. If we pick out 3 even numbers and 0 odd numbers, this can be done in

$\dbinom 33 \dbinom 30 = 1\cdot1 = 1$

way.

The total count is then the sum of all possible selections with at least 1 even number and between 0 and 3 odd numbers.

$\displaystyle \sum_{e=1}^3 \binom 3e \sum_{o=0}^3 \binom 3o = 2^3 \sum_{e=1}^3 \binom3e = 8 \left(\sum_{e=0}^3 \binom3e - \binom30\right) = 8(2^3 - 1) = \boxed{56}$

where we use the binomial identity

$\displaystyle \sum_{k=0}^n \binom nk = \sum_{k=0}^n \binom nk 1^{n-k} 1^k = (1+1)^n = 2^n$

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1 year ago