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3 years ago

Find the product of 4 2/7 x 3 1/2

Mathematics

2 answers:

Fiesta28 [93]3 years ago

8 0

Answer:

Hey there!

The product of these two fractions would equal 15.

Hope this helps :)

Delicious77 [7]3 years ago

3 0

Answer:

Hi! The answer to your question is 7 11/14 or rounded will be 15

Step-by-step explanation:

So first let’s take the whole numbers which is 4 and 3 if we add them up we will get 7.

Now we do LCD (least common denominator)

4/14+7/14=11/14

So the answer is 7 11/14 or 15

(In mixed number the answer is 7 11/14 and in whole the answer is 15)

Hope this helps! :)

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Hot dogs at a carnival cost 2.99 each plus 7%tax.what is the total cost for one hot dog?

Mars2501 [29]

$3.20
2.99 • 7% = 0.2093
2.99 + 0.2093 = 3.20

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2 years ago

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miskamm [114]

Answer:

Step-by-step explanation:

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2 years ago

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It took Aiden 6 hours to plant 8 trees. How many hours would it take for him to plant 20 trees? 24, 18, 23, 15

Pepsi [2]

<h3>Answer: 15 hours (choice D)</h3>

================================================

Work Shown:

(6 hours)/(8 trees) = (x hours)/(20 trees)

6/8 = x/20

6*20 = 8*x ... cross multiply

120 = 8x

8x = 120

x = 120/8

x = 15

It takes him 15 hours to plant 20 trees.

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2 years ago

Consider the equation below.

Korvikt [17]

Answer:

Equation in square form:

$y=3(x+5)^2-4$

Extreme value:

$(h,k)=(-5,-4)$

Step-by-step explanation:

We are given

$y=3x^2+30x+71$

we can complete square

$y=3(x^2+10x)+71$

we can use formula

$a^2+2ab+b^2=(a+b)^2$

$y=3(x^2+2\times x\times 5)+71$

now, we can add and subtract 5^2

$y=3(x^2+2\times x\times 5+5^2-5^2)+71$

$y=3(x^2+2\times x\times 5+5^2)-3\times 5^2+71$

$y=3(x+5)^2-75+71$

So, we get equation as

$y=3(x+5)^2-4$

Extreme values:

we know that this parabola

and vertex of parabola always at extreme values

so, we can compare it with

$y=a(x-h)^2+k$

where

vertex=(h,k)

now, we can compare and find h and k

$y=3(x+5)^2-4$

we get

h=-5

k=-4

so, extreme value of this equation is

$(h,k)=(-5,-4)$

6 0

3 years ago

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Before sending track and field athletes to the Olympics, the U.S. Holds a qualifying meet. The box plots below show the distance

sleet_krkn [62]

Question:

The options are;

A. The distances in the Olympic final were farther on average.

B. The distances in the Olympic final varied noticeably more than the US qualifier distances

C. The distances in the Olympic final were all greater than the US qualifier distances

D. none of the above

Answer:

The correct option is;

A. The distances in the Olympic final were farther on average.

Step-by-step explanation:

From the options given, we have

A. The distances in the Olympic final were farther on average.

This is true as the sum of the 5 points divided by 5 is more in the Olympic final

B. The distances in the Olympic final varied noticeably more than the US qualifier distances

This is not correct as the difference between the upper and lower quartile in the Olympic final is lesser than in the qualifier

C. The distances in the Olympic final were all greater than the US qualifier distances

This is not correct as the max of the qualifier is more than the lower quartile in the Olympic final

D. none of the above

We have seen a possible correct option in option A

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2 years ago

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