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3 years ago

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Maya has 5 over 6 cup of ice cream. How many 1 over 4-cup servings are in 5 over 6 cup of ice cream? ASAP I need help

1 answer:

Ierofanga [76]3 years ago

8 0

Answer:

Find the LCD (least common denominator)

{5}/{6} and {1}/{4}

it would be 12 (6x2=12 and 4x3=12)

so you multiply {5}/{6} by 2 and get {10}/{12}

then multiply {1}/{4} by 3 and get {3}/{12}

you get 3 WHOLE servings from {5}/{6} cups of icecream

Step-by-step explanation:

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The end points of one diagonal of a rhombus are (-5,2) and (1,6). If the coordinates of the 3rd vertexare (-6,10), what are the

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Look at the picture.

$E\left(\dfrac{-5+1}{2};\ \dfrac{2+6}{2}\right)\to E\left(\dfrac{-4}{2};\ \dfrac{8}{2}\right)\to E(-2;\ 4)$

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$D(x;\ y)\\\\\vec{ED}=[x-(-2);\ y-4]=[x+2;\ y-4]\\\\\vec{ED}=\vec{BE}\ therefore\\\\ \ [x+2;\ y-4]=[4;\ -6]\to x+2=4\ and\ y-4=-6\\\\x=2\ and\ y=-2\\\\Answer:\ \boxed{A.\ (2;\ -2)}$

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3 years ago

Find the median of the data.

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92.5

Step-by-step explanation:

<u>The data given:</u>

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2 years ago

Help me solve C=AB+D for B.

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3 years ago

Find the approximate side length of a square game board with an area of 123 inches to the second power

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3 years ago

Simplify the expression state any excluded values<br> 2a^2-4a+2<br> ---------------<br> 3a^2-3

chubhunter [2.5K]

Answer:

The simplified form is $\dfrac{2(x-1)}{3(x+1)}$ .

$x =1$ is the excluded value for the given expression.

Step-by-step explanation:

Given:

The expression given is:

$\dfrac{2a^2-4a+2}{3a^2-3}$

Let us simplify the numerator and denominator separately.

The numerator is given as $2a^2-4a+2$

2 is a common factor in all the three terms. So, we factor it out. This gives,

$=2(a^2-2a+1)$

Now, $a^2-2a+1=(a-1)(a-1)$

Therefore, the numerator becomes $2(a-1)(a-1)$

The denominator is given as: $3a^2-3$

Factoring out 3, we get

$3(a^2-1)$

Now, $a^2-1$ is of the form $a^2-b^2=(a-b)(a+b)$

So, $a^2-1=(a-1)(a+1)$

Therefore, the denominator becomes $3(a-1)(a+1)$

Now, the given expression is simplified to:

$\frac{2a^2-4a+2}{3a^2-3}=\frac{2(x-1)(x-1)}{3(x-1)(x+1)}$

There is $(x-1)$ in the numerator and denominator. We can cancel them only if $x\ne1$ as for $x=1$ , the given expression is undefined.

Now, cancelling the like terms considering $x\ne1$ , we get:

$\dfrac{2a^2-4a+2}{3a^2-3}=\dfrac{2(x-1)}{3(x+1)}$

Therefore, the simplified form is $\dfrac{2(x-1)}{3(x+1)}$

The simplification is true only if $x\ne1$ . So, $x =1$ is the excluded value for the given expression.

8 0

3 years ago