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1 year ago

If 6 burgers cost $7.74, how many can you buy for $19.35? Pls show work and explain how you did it :)

Mathematics

2 answers:

dimulka [17.4K]1 year ago

6 0

Answer: 15 burgers

Step-by-step explanation:

7.74/6=$1.29 per burger

$19.35* 1 burger/$1.29=

19.35*1/1.29=

19.35/1.29=15 burgers

Ilia_Sergeevich [38]1 year ago

6 0

For this problem you need to set it up in a certain way. To solve this problem you need to do the butterfly method, as shown in the picture I provided.
It doesn’t matter what goes on top or bottom as long as they are the same. For example the cost will remain the numerator all the way through and the burgers will be the denominator.
Everything else is shown in the image, hopes this helps!

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45 POINTS PLEASE HELP question 1) Given: f(x) = 3 - x; g(x) = -2x Find g[f(-1)].

andrew11 [14]

1) Given: f(x) = 3 - x; g(x) = -2x Find g[f(-1)].

g(f(x)) = -2(3 - x) = 2x - 6

g(f(-1)) = 2(-1) - 6 = -2 -6 = -8

g(f(-1)) = -8

Looks like the question 2 and 1 are the same

Both find g(f(-1))

5 0

4 years ago

Two sides of a triangle are 8cm and 12cm. Which of the following CANNOT be the measure of the third side,

blondinia [14]

Answer:

4 cannot be the measure of the third side. This is because of the Triangle Inequality Theorem, which states that the sum of two sides of a triangle must be greater than the third side (A+B>C, A+C>B, B+C>A) In this example, if side C were 4, side C (4) plus side A (8) would be 12. Since side B is 12, and 12 cannot be greater than 12, 4 would not work.

Answer=8

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

Answer:

a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

7 0

3 years ago

A 12-oz box of cereal costs $1.68. What is the cost per Ounce?

mixas84 [53]

It’s should be $0.14 if I did the math correctly

5 0

3 years ago

Can someone please help me , I really need it

Nikitich [7]

((1/3 + 1)/(3))/(4)
((1/3 + 1)/(3))/(4) = (1/3 + 1)/(3×4):
(1/3 + 1)/(3×4)
Put 1 + 1/3 over the common denominator 3. 1 + 1/3 = 3/3 + 1/3:
(3/3 + 1/3)/(3×4)
3/3 + 1/3 = (3 + 1)/3:
((3 + 1)/3)/(3×4)
3 + 1 = 4:
(4/3)/(3×4)
4/(3×3×4) = 4/(3×4×3):
4/(3×4×3)
4/(3×4×3) = 4/4×1/(3×3) = 1/(3×3):
1/(3×3)
3×3 = 9:
Answer: 1/9 gallons of yellow paint

((1/4 + 1)/(4))/(3)
Put 1 + 1/4 over the common denominator 4. 1 + 1/4 = 4/4 + 1/4:
((4/4 + 1/4)/(4))/(3)
4/4 + 1/4 = (4 + 1)/4:
(((4 + 1)/4)/(4))/(3)
4 + 1 = 5:
((5/4)/4)/3
4×4 = 16:
5/(16×3)
16×3 = 48:
Answer: 5/48 Gallons of green paint

(7/3×1/4)/8
(7/3×1/4)/8 = 7/(3×4×8):
7/(3×4×8)
3×4 = 12:
7/(12×8)
12×8 = 96:
Answer: 7/96 Gallons of blue paint

Combine:

1/9 + 5/48 + 7/96
Put 1/9 + 5/48 + 7/96 over the common denominator 288. 1/9 + 5/48 + 7/96 = 32/288 + (6×5)/288 + (3×7)/288:
32/288 + (6×5)/288 + (3×7)/288
6×5 = 30:
32/288 + 30/288 + (3×7)/288
3×7 = 21:
32/288 + 30/288 + 21/288
32/288 + 30/288 + 21/288 = (32 + 30 + 21)/288:
(32 + 30 + 21)/288
| 3 | 2
| 3 | 0
+ | 2 | 1
| 8 | 3:Answer: 83/288 or 0.2881944... Gallons are left over

6 0

3 years ago