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otez555 [7]
3 years ago
9

Victoria buys 30 ounces of gummy bears and 47 ounces of skittles. What is the ratio of skittles to gummy bears?

Mathematics
1 answer:
earnstyle [38]3 years ago
8 0

Hi there! :D The answer to your question is 47/30. The way I got it is because ratio means fractions or part of so it says skittles first so skittles go first and gummy bears go on the bottom so it is 47/30. Yummy! ;)

Hope I helped! =D


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21) What is the slope of the perpendicular lines?<br> y = 5x
Pachacha [2.7K]
The slope of the perpendicular line of y=5x would be -1/5, since slopes are always opposite reciprocals when they are perpendicular to each other.
8 0
3 years ago
Slope-intercept form of y + 3 = 4(x – 5).
Nataly [62]

Answer:

y =  − 4 x + 23

Hope this helps

8 0
3 years ago
Plz help!! Dont answer if you cant help
Nezavi [6.7K]

Answer:

94.88 cubic units

Step-by-step explanation:

Volume of the prism = Area of base * Height

= 5( \frac{1}{2}  \times 2.75 \times 4) \times 3.45 \\  \\  = 5 \times 2.75 \times 2 \times 3.45 \\  \\  = 10\times 2.75  \times 3.45 \\  \\  = 94.875 \\  \\   \approx 94.88 \: cubic \: units

7 0
3 years ago
What's the ratio 48:30
Vladimir79 [104]
24:15 would be the simplified version, if that's what you meant.
7 0
3 years ago
Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by
11111nata11111 [884]

Answer:

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is V=l\times b\times h

V=(11-2x)\times (7-2x)\times x

V=4x^3-36x^2+77x

Derivate w.r.t x,

V'(x)=4(3x^2)-2(36x)+77

V'(x)=12x^2-72x+77

The critical point when V'(x)=0

12x^2-72x+77=0

Solve by quadratic formula,

x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}

x=4.607,1.392

Derivate again w.r.t x,

V''(x)=24x-72

Now, V''(4.607)=24(4.607)-72=38.568>0 (+ve)

V''(1.392)=24(1.392)-72=-38.592 (-ve)

So, there is maximum at x=1.392.

The length of the box is l=11-2x

l=11-2(1.392)=8.216

The breadth of the box is b=7-2x

b=7-2(1.392)=4.216

The height of the box is h=1.392.

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is V=l\times b\times h

V=8.216\times 4.216\times 1.392

V=48.217\ in.^3

The volume of the box is 8.217 cubic inches.

5 0
3 years ago
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