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3 years ago

PLS HELP WILL GIVE BRAINLEST AND THANKS!!!!!!!!!!!

Mathematics

2 answers:

Stels [109]3 years ago

8 0

Answer: 2 1/4 more flour for 9 servings

Step-by-step explanation: (a) how many total cups of flour are there per serving? show your work

1 1/2÷6= 3/2 x 1/6 = 3/12= 1/4 cups of flour per serving

(b) how many total cups of sugar (white and brown) are there per serving?show your work

3/4 + 1/3 = 9/12 + 4/12= 13/12= 1 1/12 total cups of sugar

13/12 ÷ 6/1 = 13/12 x1/6 = 13/72 per serving

(c) supposed you modify the recipe so that it makes 9 servings. How much more flour do you need for the modified recipe than you need for the original recipe?Show your work/

1/4 x 9 = 9/4 = 2 1/4 more flour for 9 servings.

kenny6666 [7]3 years ago

6 0

Answer:

a) 0.25 cups of flour

(b) 0.18 cups of sugar

Step-by-step explanation:

The recipe requires 3 amounts of ingredients to make 6 servings. The following is needed:

1. 1 1/2 cups of flour

2. 3/4 cups of white sugar

3. 1/3 cups of brown suger

a) The total cups of flour per serving can be determined by dividing the flour needed for 6 servings by 6:

= 1.5/6=0.25

0.25 cups of flour is needed per serving

b) The total cups of brown and white sugar can be determined by divided both quantites by 6 servings.

For brown sugar = 0.75/6 = 0.125 cups of brown sugar

For white sugar = 0.333/6= 0.0555≈0.055 cups of white sugar

Total amount of sugar = brown sugar + white sugar = 0.125+0.055=0.18 cups of sugar

c) If we can make 9 servings we can determine the the percentage of more servings and the amount of flour for the modified recipe.

The percentage change of recipes = ((9-6)/9)*100=33.33%

Therefore the new amount of flour needed is:

1.5x0.3333+1.2=0.2

Therefore we need 2-1.5=0.5 cups more flour than the original recipe

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Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Shalnov [3]

Hey there, hope I can help!

$\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}$
$2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)$

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
$x^2-2x-46=0$

Lets use the quadratic formula now
$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$
$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}$

$\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

Multiply the numbers 2 * 1 = 2
$\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ \sqrt{4+184} \ \textgreater \ \sqrt{188} \ \textgreater \ 2 + \sqrt{188}$
$\frac{2+\sqrt{188}}{2} \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ \sqrt{2^2\cdot \:47}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}$

$Factor\;2+2\sqrt{47} \ \textgreater \ Rewrite\;as\;1\cdot \:2+2\sqrt{47}$
$\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+\sqrt{47}\right) \ \textgreater \ \frac{2\left(1+\sqrt{47}\right)}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ 1+\sqrt{47}$

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

$\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

$\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ 2-\sqrt{188} \ \textgreater \ \frac{2-\sqrt{188}}{2}$

$\sqrt{188} = 2\sqrt{47} \ \textgreater \ \frac{2-2\sqrt{47}}{2}$

$2-2\sqrt{47} \ \textgreater \ 2\left(1-\sqrt{47}\right) \ \textgreater \ \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \ 1-\sqrt{47}$

Therefore our final solutions are
$x=1+\sqrt{47},\:x=1-\sqrt{47}$

Hope this helps!

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