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3 years ago

HELP MEEEEEEEEEEEEEEEEEEEEEEEEEEE

Mathematics

2 answers:

Marizza181 [45]3 years ago

4 0

Answer:

A. 3 batches

B. 180 cookies

Step-by-step explanation:

Marysya12 [62]3 years ago

4 0

Answer:

A = 3 B = 180

Step-by-step explanation:

First you do 21/4 divided by 7/4 = 3. So 3 equals how many batches. Now since 5 dozen cookies in each batch and a dozen equals 12 cookies there are 60 cookies in each batch and since there are 3 batches you would do 60 * 3 = 180.

You might be interested in

Simplify the cube root in simplest radical form.<br> 9x374<br> Please help

algol [13]

Answer:

D. $xy\sqrt[3]{9y}$

Step-by-step explanation:

$\sqrt[3]{9x^3y^4}$

$\sqrt[3]{9}\sqrt[3]{x^3}\sqrt[3]{y^4}$

The $\sqrt[3]{x^3}$ cancels out to become x:

$\sqrt[3]{9}x\sqrt[3]{y^4}$

Split the $y^4=y*y^3$

$\sqrt[3]{9}x\sqrt[3]{y^3}\sqrt[3]{y^1}$

$\sqrt[3]{y^3} =y$

$xy\sqrt[3]{9} \sqrt[3]{y}$

Put the cube root of y and cube root of 9 together:

$xy\sqrt[3]{9y}$

6 0

3 years ago

One of our brainliest, Konrad509, made this:

Reil [10]

$\dfrac{B_x \sqrt{74_x}}{1D_x}+J_x51_x=4G3_x$

A=10, B=11, C=12, etc.

$\dfrac{11\cdot x^0\cdot \sqrt{7\cdot x^1+4\cdot x^0}}{1\cdot x^1+13\cdot x^0}+19\cdot x^0\cdot (5\cdot x^1+1\cdot x^0)=4\cdot x^2+16\cdot x^1+3\cdot x^0\\\\\dfrac{11\sqrt{7x+4}}{x+13}+19(5x+1)=4x^2+16x+3\\\\\dfrac{11\sqrt{7x+4}}{x+13}+95x+19=4x^2+16x+3\\\\11\sqrt{7x+4}+95x(x+13)+19(x+13)=(4x^2+16x+3)(x+13)\\\\11\sqrt{7x+4}+95x^2+1235x+19x+247=4x^3+52x^2+16x^2+208x+3x+39\\\\11\sqrt{7x+4}=4x^3-27x^2-1043x-208\\\\121(7x+4)=(4x^3-27x^2-1043x-208)^2$

$121(7x+4)=(4x^3-27x^2-1043x-208)^2\\\\847x+484=16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433888 x + 43264\\\\16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433041 x +42780=0$

Now, the "only" thing that remains to do is solving the above equation.

While making this problem I only made sure it has a solution. I didn't try to solve it myself and I didn't know it will end up with such "convoluted" polynomial. Sorry to everyone who tried to solve it... m(_ _)m

I think the best way to approach it is using the rational root theorem since we know that $x\in\mathbb{N}$ . Moreover we can deduce that $x\geq19$ since there is $J$ and $J=19$ .