Ask question

Ask question

All categories

2 years ago

11

One box of crackers costs $1.75. The crackers are advertised as “3 boxes for $5.25.” Which proportion can be used to represent t

he cost of the crackers?

2 answers:

Art [367]2 years ago

6 0

Answer:

$\frac{1}{3}=0.33$

Step-by-step explanation:

Given : One box of crackers costs $1.75. The crackers are advertised as “3 boxes for $5.25.”

To find : Which proportion can be used to represent the cost of the crackers?

Solution :

The cost of 1 box is = $1.75

The cost of 3 boxes is = $5.25

A proportion is when two ratios are equal.

Ratio of boxes is equal to the ratio of cost of crackers.

Therefore, 1 : 3 = 1.75 : 5.25

In Proportion it can be written as $\frac{1}{3}=\frac{1.75}{5.25}$

This is true also, as $\frac{1}{3}=0.33$

$\frac{1.75}{5.25}=0.33$

Daniel [21]2 years ago

5 0

A. 1/1.75 = 3/5.25

Ignore this part I'm just trying to get at least 20 characters.

You might be interested in

(c) if the corresponding terms of two arithmetic sequences are added, is the resulting sequence arithmetic?

Harman [31]

To determine whether the corresponding terms of 2 arithmetic sequence's added will give new arithmetic sequence or not, Let' take 2 Arithmetic sequences.

In one first term is a1 and common difference is d1, in the other first term is a2 and common difference is d2.

Now nth term for first sequence = a1+(n-1) d1

nth term for second sequence = a2+(n-1) d2

Now add the 2 terms: a1+(n-1)d1 +a2 +(n-1)d2

= a1+a2 + (n-1)(d1+d2)

This is again new arithmetic sequence with first term a1+a2 and common difference d1+d2.

Hence if we add corresponding terms of 2 arithmetic sequence, we will again get an arithmetic sequence.

5 0

1 year ago

A cardboard popcorn container is shaped like a cone that is 11 inches high and has a radius of 2.5 inches. What is the volume of

Vika [28.1K]

Vcone=(1/3)hpir^2
h=11
r=2.5
Vcone=(1/3)11pi2.5^2
Vcone=(11/3)pi6.26
Vcone=68.75/3*3.14
Vcone=71.95833
round
Vcone=71.96

vcone=71.95 in^3

8 0

2 years ago

A delivery truck is stocked with boxes of leather footballs. The graph shows the relationship between the number of footballs an

pashok25 [27]

Answer:

21

Step-by-step explanation:

To find the answer, you'd have to continue tracing the diagonal line until it intersects with the vertical line that corresponds to the number 3.

If you do that, you'll see that in3 boxes, there are 21 footbals.

You can also calculate it mathematically:

If you have 7 balls per box as shown in the graph, you just have to multiply 7 by 3 to know how many balls you'll find in 3 boxes. 7 * 3 = 21.

Hope it helped,

BioTeacher101

6 0

1 year ago

Read 2 more answers

SOMEEEEONEEE pls help pls and thx :)

alekssr [168]

1) The sculptor created a marble basin with an approximate volume of 94,509.6 cubic centimeters - Option D, 2) The approximate area of the heart-shaped cake is 283 square inches - Option B, 3) The length of the wooden frame around the window is 94.3 inches - Option B.

In this question we should make use of geometric formulas for volumes and areas and key information from statement in order to find the right choices.

1) In this case, the volume occupied by the marble water basin ( $V$ ), in cubic centimeters, by subtracting the volume of the hemisphere from the volume of the cylinder:

$V = \pi\cdot R^{2}\cdot h - \frac{2\pi}{3} \cdot r^{3}$ (1)

Where:

$R$ - Radius of the cylinder, in centimeters.
$h$ - Height of the cylinder, in centimeters.
$r$ - Radius of the cylinder, in centimeters.

If we know that $R = 30\,cm$ , $h = 45\,cm$ and $r = 25\,cm$ , then the volume of the marble is:

$V = \pi \cdot (30\,cm)^{2}\cdot (45\,cm) - \frac{2\pi}{3}\cdot (25\,cm)^{3}$

$V \approx 94509.579\,cm^{3}$

The right choice is D.

2) To determine the approximate surface area of the cake covered in red frosting ( $A_{s}$ ), in square inches, we need to find the sum of the surface area of the entire circle and the surface area of the square:

$A_{s} = l^{2} + 2\cdot l\cdot h + \frac{\pi}{4}\cdot D^2 + \pi \cdot D\cdot h$ (2)

Where:

$l$ - Side length of the square, in inches.
$h$ - Height of the cakes, in inches.
$D$ - Diameter of the cakes, in inches.

If we know that $l = 9\,in$ , $h = 3\,in$ and $D = 9\,in$ , then the <em>approximate</em> area covered in red frosting:

$A_{s} = (9\,in)^{2} + 2\cdot (9\,in)\cdot (3\,in) + \frac{\pi}{4}\cdot (9\,in)^{2} + \pi \cdot (9\,in)\cdot (3\,in)$

$A_{s} \approx 283.440\,in^{2}$

The right choice is B.

3) The frame around the window is found by means of the following perimeter formula ( $s$ ), in inches:

$s = \pi\cdot r + 2\cdot h + 2\cdot r$ (3)

Where:

$r$ - Radius, in inches.
$h$ - Height of the rectangle, in inches.

If we know that $r = 9\,in$ and $h = 24\,in$ , then the length of the frame around the window is:

$s = \pi\cdot (9\,in) + 2\cdot (24\,in) + 2\cdot (9\,in)$

$s \approx 94.274\,in$

The right choice is B.

We kindly invite to see question on volumes: brainly.com/question/1578538

7 0

1 year ago

Given: ∠BCD is right; BC ≅ DC; DF ≅ BF; FA ≅ FE Triangles A C D and E C B overlap and intersect at point F. Point B of triangle

Aloiza [94]

Answer:

1) ΔCBF ≅ ΔCDF by (SSS)

2) ΔBFA ≅ ΔDFE by (SAS)

3) ΔCBE ≅ ΔCDA by (HL)

Step-by-step explanation:

1) Since BC ≅ DC and DF ≅ BF where CF ≅ CF (reflective property) we have;

ΔCBF ≅ ΔCDF by Side Side Side (SSS) rule of congruency

2) Since DF ≅ BF and FA ≅ FE where ∠DFE = ∠BFA (alternate angles)

Therefore;

ΔBFA ≅ ΔDFE by Side Angle Side (SAS) rule of congruency

3) Since FA ≅ FE and DF ≅ BF then where EB = FE + BF and AD = FA + DF

Where:

EB and AD are the hypotenuse sides of ΔCBE and ΔCDA respectively

We have that;

EB = AD from FE + BF = FA + DF

Where we also have BC ≅ DC

Where:

BC and DC are the legs of ΔCBE and ΔCDA respectively

Then we have the following relation;

ΔCBE ≅ ΔCDA by Hypotenuse Leg (HL).

4 0

1 year ago

Read 2 more answers