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docker41 [41]
4 years ago
9

A recipe for sabayon calls for 2 egg yolks, 3 tablespoons of sugar, and 1⁄4 cup of white wine. After cracking the eggs, you star

t measuring the sugar, but accidentally put in 4 tablespoons of sugar. How can you compensate?
Mathematics
1 answer:
Annette [7]4 years ago
3 0

Answer:

Step-by-step explanation:

A recipe for sabayon calls for 2 egg yolks, 3 tablespoons of sugar, and ¼ cup of white wine. After cracking the eggs, you start measuring the sugar, but accidentally put in 4 tablespoons of sugar. How can you compensate?

Note that we put  in 4/3 more sugar than we wanted, because 3(4/3) = 4

So we need to increase eveything else by 4/3

So 2(4/3)  = 8/3  = 2 +2/3 egg yolk    and      (1/4)(4/3) =4/12 = 1/3 cup of wine

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Select the correct answer
tangare [24]

Answer: E.  (x+5)^2 + (y-5)^2 = 9

Step-by-step explanation:

The equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius.

We are given that the center is (-5,5), which means that h = -5 and k = 5.

We are also given that the radius (r) = 3, which means r^2 = 9.

Therefore, the equation should be (x-(-5))^2 + (y-5)^2 = 9.

--> (x+5)^2 + (y-5)^2 = 9

3 0
2 years ago
Starting at home, Luis traveled uphill to the gift store for 50 minutes at just 6 mph. He then traveled back home along the same
Georgia [21]

Answer:

Therefore his average speed for the entire trip is 8 mph.

Step-by-step explanation:

Given, In 50 minutes Luis traveled uphill to gift store at 6 mph.

6 mph  means in 1 h= 60 minutes Luis can covered 6 mile.

In 1 minutes Luis can covered \frac{6}{60} mile.

In 50 minutes Luis can covered \frac{6\times 50}{60}  mile

                                                  = 5 mile

Again when he come back at home, the speed was 12 mph.

12 mph  means in 1 h= 60 minutes Luis can covered 12 mile.

Therefore he traveled 12 mile in 60 minutes

He traveled 1 mile in \frac{60}{12}  minutes.

He traveled 5 mile in \frac{60\times 5}{12} minutes

                                   = 25 minutes.

Total distance for the entire trip is =(5+5) mile=10 mile

Total time for the entire trip is = (50+25) minutes = 75 minutes

\textrm{Average speed}=\frac{\textrm{Total distance}}{\Textrm {total time}}

                       =\frac{10}{75} m/min

                      =\frac{10\times 60}{75} mph

                      = 8 mph

Therefore his average speed for the entire trip is 8 mph.

4 0
3 years ago
Claire purchases a new dress for the prom. The dress is priced $160, but it is on sale for 30% off. Claire’s aunt works at the s
stiks02 [169]
The answer should be c. 108.36
4 0
4 years ago
Read 2 more answers
Explain, in complete sentences, which method you would use to solve the following system of equations and why you chose that met
svp [43]
I would use elimination. I would multiply 4x+3y=2 by 3 so the it becomes 12x+9y=6, and I would multiply the bottom equation by 4 so that it becomes 12x+8y=4. Because I am using elimination I have to get rid of one variable, I chose to get rid of X hence that the coefficient of X in each equation is now 12. i simply subtract both equations and get y by itself. y=2. I plug in Y to find x in either of the original equations, where I get x=-1. To check my work substitute the values of x and y in the other equation. 4(-1)+3(2)=2 ------->  2=2
4 0
3 years ago
Hey! Can someone check my answers?
Andrew [12]
The greatest common factor is the biggest number taken from the values.

Q1. The answer is <span>A. 5y^6
</span>
20 y^{9} +5 y^{6}= 4*5 y^{9}+5 y^{6}
Since x^{a}* x^{b}  =x^{a+b}, then x^{9}= x^{3}* x^{6}

Back to our expression:
4*5 y^{9}+5 y^{6}=4*5 y^{3}*y^{6}+5 y^{6}=4 y^{3}*5y ^{6}+  5y ^{6}*1=5 y^{6} (4y ^{3} +1)
The greatest common factor is thus 5 y^{6}


Q2. The answer is <span>D. 12xy^2
</span>
12x y^{5}+60 x^{4} y^{2} -24 x^{3} y^{3}=12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}
We will use the rule  x^{a}* x^{b} =x^{a+b} to factorise the exponents:
12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}= \\ =12x*y^{2}*y^{3}+5*12*x* x^{3} *y^{2}-2*12x* x^{2} *y*y^{2}= \\ =12xy^{2}*y^{3}+12xy^{2}*5 x^{3} -12xy^{2}*2 x^{2} y= \\ =12xy^{2}(y^{3}+5 x^{3}-2 x^{2} y)
The greatest common factor is thus 12xy^{2}
3 0
3 years ago
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