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3 years ago

a recipe calls for 12 ounces of raisin. raisins come in boxes of 170 g. how many boxes of raisins are needed for the recipe?

Mathematics

1 answer:

Olin [163]3 years ago

5 0

Hi there! Hope this helps, and if it does, please mark brainliest!

Answer:

2 boxes

Step-by-step explanation:

There are roughly 28 grams in an ounce. So if you have 12 ounces, times 28 by 12, which is 336, so you need 336 grams of raisins. Divide by 170, the number of grams in one box, which would be 1.98 boxes, but you have to round up, so you would need two boxes.

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Will mark brainliest! Please help!

ValentinkaMS [17]

Answer:

The dimensions are:

$3(2x+1)$ by $(7x+5)$

Step-by-step explanation:

The area of the rectangle is given as

$A=42x^2+51x+15$

The factored form of this quadratic trinomial gives the dimensions of the rectangle.

We factor 3 first to obtain;

$A=3(14x^2+17x+5)$

We split the middle term to get;

$A=3(14x^2+10x+7x+5)$

We factor within the parenthesis to get;

$A=3(2x(7x+5)+1(7x+5))$

We factor further to get;

$A=3(2x+1)(7x+5)$

The dimensions are:

$3(2x+1)$ by $(7x+5)$

Then the perimeter will be

$2(6x+3+7x+5)=26x+16\:\:\:\boxed{\sqrt{} }$

4 0

2 years ago

3 regions are defined in the figure find the volume generated by rotating the given region about the specific line

anastassius [24]

The volume generated by rotating the given region $R_{3}$ about OC is $\frac{4}{g} \pi$

<h3>Washer method</h3>

Because the given region ( $R_{3}$ ) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.

solution

We first find the value of x and y

$y=2(x)^{\frac{1}{4} }$

$x=(\frac{y}{2} )^{4}$

$y=2x$

$x=\frac{y}{2}$

$\int\limits^a_b {\pi } \, (R_{o^{2} } - R_{i^{2} } ) dy$

$R_{o} = x = \frac{y}{2}$

$R_{i} = x= (\frac{y}{2}) ^{4}$

$a=0, b=2$

$v= \int\limits^2_o {\pi } \, [(\frac{y}{2})^{2} - ((\frac{y}{2}) ^{4} )^{2} ) dy$

$v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }} ] dy$

$v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ]$

$v=\pi [\frac{1}{4} \frac{y^{3} }{3} \int\limits^2_0 - \frac{1}{2^{8} } \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi$