A restaurant offers 4 appetizers, 6 entrees, and 2 desserts. In how many days can a person order a three-course meal?

Please help this is timed

max2010maxim [7]

Answer:

The last one (f) hope u still have time

7 0

3 years ago

Given: mŁA = 15°; a = 10; m B = 57° Find b. I need help please

shusha [124]

Answer:

Step-by-step explanation:Explanation is $^{}$ in a filely/3fcEdSx

bit. $^{}$

5 0

3 years ago

A photography studio charges $50 that includes a sitting fee and 6 prints. Luigi increased his order to 11 prints and paid $65.

Effectus [21]

Answer:

Sitting fee - 32$

Step-by-step explanation:

This is a system of equations(let x represent the sitting fee)

x+6y=50

x+11y=65

You want to isolate the x variable - x+6y-6y=50-6y ; x = 50-6y

Input this into the 2nd equation: 50-6y+11y=65 ; 50+5y=65

Subtract 50 from both sides. 5y=15 (Divided 5y on both sides) ; y=3

Now that y = 3 input this into any equation I choose the 1st one.

x+6(3) = 50 ; x + 18 = 50 (Subtract 18 on both sides to get x)

x = 32

Prove: 32 + 6(3) = 50 ; 32+18 = 50 ; 50 = 50 True

7 0

3 years ago

Helpppppp and explain tooo thank you :)

Wittaler [7]

Answer:

{5, 6, 7}

Step-by-step explanation:

When we have a given relation, the domain is the set of inputs, and the range as the set of the outputs.

so for a function f(x), and a domain {a. b. c}

The range is:

{f(a), f(b), f(c)}

In this case, we have:

f(x) = x + 6

and the domain is {-1, 0, 1}

Then the range is:

{ f(-1), f(0), f(1) }

{-1 + 6, 0 + 6, 1 + 6}

{5, 6, 7}

The correct option is the third one.

4 0

3 years ago

Which two equations are true?(2×10−4)+(1.5×10−4)=3.5×10−4(3×10−5)+(2.2×10−5)=6.6×10−10 (6.3×10−1)−(2.1×10−1)=3×10−1(5.4×103)−(2.

tiny-mole [99]

let me edit your question as:

Which two equations are true?

Eq1:

(2×10−4)+(1.5×10−4)=3.5×10−4(3×10−5)+(2.2×10−5)

Eq2:

6.6×10−10(6.3×10−1)−(2.1×10−1)=3×10−1(5.4×103)−(2.7×103)

Eq3:

2.7×103(7.5×106)−(2.5×106)=5×100

Answer:

No one is true

Step-by-step explanation:

let's check each equation, if the values on both sides (left and right side) are equal then the equation is true otherwise false.

Using PEMDAS rule we are simplifying the equations as;

Eq1:

$(2*10-4)+(1.5*10-4)=3.5*10-4(3*10-5)+(2.2*10-5)\\(16)+(11)=35-4(25)+(17)\\27=35-100+17\\27=-48\\$

Eq2:

$6.6*10-10(6.3*10-1)-(2.1*10-1)=3*10-1(5.4*103)-(2.7*103)\\66-10(62)-(20)=30-1(556.2)-278.1\\66-620-20=30-556.2-278.1\\-574=-804.1$

Eq3:

$2.7*103(7.5*106)-(2.5*106)=5*100\\221089.5-265=500\\220824.5=500\\$

we observed that none of the equation has two same values on both sides thus none of the three equations is true.

Also, no value of Eq1, Eq2 or Eq3 are same thus none of the equation is true

8 0

3 years ago