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3 years ago

Ariana made cookies. She used 3 7/8 cups of flour and 2 1/8 cups of sugar. How much more flour than sugar did Ariana use?

Mathematics

2 answers:

Vika [28.1K]3 years ago

7 0

Answer:

1 6/8 cups of flour

Step-by-step explanation:

Just subtract 2 1/8 from 3 7/8

tino4ka555 [31]3 years ago

3 0

1 6/8 more flour than sugar

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Simplify the expression.

Tamiku [17]

Answer:

$\boxed{=144--> C.}$

Step-by-step explanation:

You had to used pemdas stands for: p-parenthesis, e-exponents, m-multiply, d-divide, a-add, and s-subtract that came from left to right.

First you calculate with parenthesis first.

$(13-1)^2/2$

$12^2/2$

Then you calculate with exponents.

$12^2=12*12=144$

$144/2$

Next, divide from left to right.

$144/2=72$

$144/72=2$

$72*2=144$

$2*72=144$

$=72$

$2*72$

Finally, multiply from left to right.

$72*2=144$

$12*12=144$

Final answer: $\boxed{=144}$

Hope this helps!

And thank you for posting your question at here on brainly, and have a great day.

-Charlie

3 0

3 years ago

Given that A, O & B lie on a straight line segment, evaluate COD.

GrogVix [38]

(3x + 94) I’m not sure

8 0

3 years ago

Read 2 more answers

Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure

Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

$\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}$

Now, we need to express the quadratic polynomials using their roots, as follows:

$ay^2+by+c=a(y-y_1)(y-y_2)$

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}$

Applying the quadratic formula to the second polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the third polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the fourth polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}$

Substituting into the rational expression and simplifying:

$\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}$

8 0

1 year ago

Help me If x=1/2 and y=2/3 , find the value of xy+ (x+ y)

Dima020 [189]

Answer:

3/2

Step-by-step explanation: