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sveta [45]
3 years ago
5

Malik is making 36 small pack of chili spices and 15 large packs of chili spices. A small pack contains 2.35 ounces of spices.A

large pack contains 10.8 ounces of spices. How many ounces of chili spices should Malik make?
Mathematics
2 answers:
Vladimir [108]3 years ago
6 0

you will need 246 ounces of chili spices

Elena L [17]3 years ago
5 0
The answer is
To find how many ounces you need to make, you multiply the number of packs by how many ounces.
So, it would look like:
36*2.35=84.6
15*10.8=162

Then you add the ounces together and you have your answer:
162+84.6=246.6 ounces
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Fiona is 7 years older than Gavin. Alex is 12 years younger than Gavin. If the total of their ages is 73, how old is the younges
sp2606 [1]

Answer:

Think that it’s Alex

7 0
2 years ago
I just need no. a please help me to prove this. ​
OleMash [197]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π    and         cos A = cos B · cos C

scratchwork:

  A + B + C = π

               A = π - (B + C)

         cos A = cos [π - (B + C)]                              Apply cos

                    = - cos (B + C)                                    Simplify

                    = -(cos B · cos C - sin B · sin C)          Sum Identity

                    = sin B · sin C - cos B · cos C               Simplify

cos B · cos C = sin B · sin C - cos B · cos C               Substitution

2cos B · cos C = sin B · sin C                                        Addition

                     2=\dfrac{\sin B\cdot \sin C}{\cos B \cdot \cos C}                                     Division

                     2 = tan B · tan C

\text{Use the Sum Identity:}\quad \tan(B+C)=\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}

<u>Proof LHS → RHS</u>

Given:                              A + B + C = π

Subtraction:                     A = π - (B + C)

Apply tan:                  tan A = tan(π - (B + C))

Simplify:                               = - tan (B + C)

\text{Sum Identity:}\qquad \qquad \qquad =-\bigg(\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}\bigg)

Substitution:                        = -(tan B + tan C)/(1 - 2)

Simplify:                               = -(tan B + tan C)/-1

                                            = tan B + tan C

LHS = RHS:   tan B + tan C = tan B + tan C  \checkmark

5 0
2 years ago
Kenji is decorating papers with a total of 15 heart stickers and 12 star stickers for the child he is babysitting. If he wants t
Marina CMI [18]

Answer: The greatest number of pages Kenji can decorate = 3

Step-by-step explanation:

Given: Total heart stickers = 15

Total star stickers =12

If all the papers identical, with the same combination of heart and star stickers and no stickers left over.

Then the greatest number of pages Kenji can decorate = GCD(15,12)  [GCD=greatest common divisor]

Since 15 = 3 x 5

12=2 x 2 x 3

GCD(15,12) =3

Hence, the greatest number of pages Kenji can decorate = 3

7 0
2 years ago
Before sending track and field athletes to the Olympics, the U.S. Holds a qualifying meet. The box plots below show the distance
sleet_krkn [62]

Question:

The options are;

A. The distances in the Olympic final were farther on average.

B. The distances in the Olympic final varied noticeably more than the US qualifier distances

C. The distances in the Olympic final were all greater than the US qualifier distances

D. none of the above

Answer:

The correct option is;

A. The distances in the Olympic final were farther on average.

Step-by-step explanation:

From the options given, we have

A. The distances in the Olympic final were farther on average.

This is true as the sum of the 5 points divided by 5 is more in the Olympic final

B. The distances in the Olympic final varied noticeably more than the US qualifier distances

This is not correct as the difference between the upper and lower quartile in the Olympic final is lesser than in the qualifier

C. The distances in the Olympic final were all greater than the US qualifier distances

This is not correct as the max of the qualifier is more than the lower quartile in the Olympic final

D. none of the above

We have seen a possible correct option in option A

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