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dusya [7]
3 years ago
10

Three ice cream cones cost $8.40. At this rate, how much do two ice cream cones cost?​

Mathematics
2 answers:
serg [7]3 years ago
6 0

Answer:

two cost 5.6 dollars

Step-by-step explanation:

8.4 / 3=2.8$

2.8*2=5.6$

snow_tiger [21]3 years ago
5 0

Answer:

$5.60

Step-by-step explanation:

You first take 8.40 and divide it by 3. Which makes you get 2.8. 2.8 is the cost per ice cream cone. Do 2.8 times 2 to get 5.6.

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If 5.5 ounces of silver is worth $80 how much could you buy with $15 (setup and solve a proportion)
Liono4ka [1.6K]

Answer:

Problem-solving requires persistence as much as it requires ingenuity. When you get stuck, or solve a problem incorrectly, back up and.

124 pages

Step-by-step explanation:

6 0
3 years ago
Two jackets have a combined cost of $98. Jacket a costs $12 less than jacket b. How much does each jacket cost?
lawyer [7]

Answer:

Jacket A costs 43$

Jacket B costs 55$

Step-by-step explanation:

According to the Question,

  • Given, Two jackets have a combined cost of $98. Jacket 'A' costs $12 less than jacket 'B'. Thus, if the jacket B Cost 'X' So, Jacket A costs 'X-12'.

Therefore, Jacket A + Jacket B = 98

X-12+X = 98

2X = 110

X = $55 ( Cost of jacket B) & Cost of Jacket A is 55-12 = 43$ .

7 0
3 years ago
An influenza outbreak occurred in a private high school dormitory that housed 83 students. Case A began on October 1 and cases B
statuscvo [17]

Answer:

The secondary attack rate is 55%

Step-by-step explanation:

Total number of students = 83

Primary cases: cases A+B+C = 3 primary cases.

Secondary cases = 44

Total cases = 44 + 3 = 47

To find the secondary attack rate, we are to assume that the primary cases are now immune because they had the illness.

We have:

Secondary attack rate = \frac{total cases - primary cases}{number of susceptible - primary cases} * 100

= \frac{47-3}{83-3} * 100

= \frac{44}{80} * 100 = 55

The secondary attack rate is 55%

8 0
3 years ago
This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex
ANTONII [103]

Answer:

Maximum value: 3* \sqrt{n}

Minimum value: -3* \sqrt{n}

Step-by-step explanation:

Let g(x) = x_1^2 + x_2^2+x_3^2+ ----+ x_n^2 , the restriction function.The Lagrange Multiplier problem states that an extreme (x1, ..., xn) of f with the constraint g(x) = 9 has to follow the following rule:

\nabla{f}(x_1, ..., x_n) = \lambda \nabla{g} (x_1,...,x_n)

for a constant \lambda .

Note that the partial derivate of f respect to any variable is 1, and the partial derivate of g respect xi is 2xi, this means that

1 = \lambda 2 x_1

Thus,

x_i = \frac{1}{2\lambda} = c

Where c is a constant that doesnt depend on i. In other words, there exists c such that (x1, x2, ..., xn) = (c,c, ..., c). Now, since g(x1, ..., xn) = 9, we have that n * c² = 9, or

c = \, ^+_- \, \sqrt{\frac{9}{n} } = \, ^+_- \frac{3}{\sqrt{n}}

When c is positive, f reaches a maximum, which is \frac{3}{\sqrt{n}}  +  \frac{3}{\sqrt{n}} +  \frac{3}{\sqrt{n}}  + ..... +  \frac{3}{\sqrt{n}}  = n *  \frac{3}{\sqrt{n}}  = 3 * \sqrt{n}

On the other hand, when c is negative, f reaches a minimum, -3 * \sqrt{n}

8 0
4 years ago
Solve by completing the square. 4x^2 + 8x– 3=0
uysha [10]

Answer:

x = -1 + \sqrt{\frac{7}{4} }, x = -1 - \sqrt{\frac{7}{4} }

Step-by-step explanation:

4x^{2} + 8x - 3 = 0

First, add 3 to both sides:

4x^{2} + 8x = 3

Next, divide both sides by 4:

x^{2} + 2x = \frac{3}{4}

Now, add the correct number by dividing 2 by 2 and squaring it and add it to both sides:

x^{2} + 2x + 1 = \frac{7}{4}

Then, factor the left side:

(x + 1)^{2} = \frac{7}{4}

Next, take the square root of both sides:

x + 1 = ± \sqrt{\frac{7}{4} }

Solve for x:

x = -1 + \sqrt{\frac{7}{4} }, x = -1 - \sqrt{\frac{7}{4} }

3 0
3 years ago
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