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vladimir1956 [14]
3 years ago
12

Tyron made 60 cakes. He gave 1/6 of the 60 cakes to Lilly. What fraction of the cakes does he have left?

Mathematics
2 answers:
Likurg_2 [28]3 years ago
8 0

Answer:

5/6

Step-by-step explanation:

Rainbow [258]3 years ago
8 0

Answer:

Step-by-step explanation:

hiiiii

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If 18% of the beads in a bag are green, 25% are red, and 40% are purple, what is the probability of choosing a green bead and th
Elina [12.6K]

the answer is a)22 %

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Plzzzzz help I don’t understand and this question is worth 23 points that’s how much I need to know how to do this and I need to
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You need PEMDAS
do 3·6 and then 12/6
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Maria mixes \frac{1}{4} cup white sugar and \frac{2}{3} cup brown sugar to make a topping for some muffins. She uses \frac{1}{12
Hoochie [10]

Answer: 11 muffins

Step-by-step explanation:

Since Maria mixes 1/4 cup white sugar with 2/3 cup brown sugar, this will give us a mixture of:

= 1/4 + 2/3

= 3/12 + 8/12

= 11/12

Ww are then told that she uses 1/12 cup of the mixture to top each muffin. Therefore, the number of muffins that she can top will be:

= 11/12 ÷ 1/12

= 11/12 × 12/1

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3 0
3 years ago
Use Cramer's Rule to solve each system.<br> 5x – 3y = 8<br> -2x - y = 10
zlopas [31]

Answer:

-2, -6

Step-by-step explanation:

using cramer rule

5    -3    

-2   -1    

calculatiing the determinant = (5 x-1) - (-3x-2) = -5 -(6) = -5-6=-11

using cramer rule

to calculate x we change the coefficient of x with the answer (8,10)

8   -3

10   -1

we calculate determinant = (8x-1)-(-3x10) = -8-(-30) = -8+30 =22

to calculate x

22/-11= -2

to calculate y we change the coefficient of y with the answer (8,10)

5   8

-2   10

we calculate determinant = (5x10)-(8x-2) = 50 -(-16) = 50 +16=66

to calculate y

66/-11= -6

6 0
3 years ago
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The following statement is either true​ (in all​ cases) or false​ (for at least one​ example). If​ false, construct a specific e
matrenka [14]

Answer:

False, counterexample below

Step-by-step explanation:

Denote the unit vectors of R^3 by e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1). Now consider v_1=e_1, v_2=2e_1 and v_3=e_3. We have that v_1, v_2,v_3 \in \mathbb{R}^3. Also, the vector v_3 is not a linear combination of v_1, v_2 because any linear combination of these two vectors will have third coordinate zero, but v_3 has third coordinate 1 so they can't be equal.

However, the set \{v_1, v_2,v_3\} is not linearly independent, because 2v_1-v_2+0v_3=0 is a non-trivial linear combination of these vectors that equals zero.

4 0
3 years ago
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