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3 years ago

The difference between sixteen, and the product of seven and two

Mathematics

2 answers:

Monica [59]3 years ago

6 0

Answer:

Step-by-step explanation:

If this product was written as an equation, it would look like this:

16 - 7 * 2

PEMDAS states that multiplication comes before addition. So, the first step in the equation is to multiply 7 and 2.

7 * 2 = 14

Now, the equation looks like:

16 - 14

All we need to do now is subtract 14 from 16.

16 - 14 = 2

So, the final answer is 2.

Andrew [12]3 years ago

5 0

16-7x2 = 2

16-7x2 (<em>multiply 7x2</em>)

16-14 (<em>subtract</em>)

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The price of<br> a watch was increased by 20% to £198.<br> What was the price before the increase?

stich3 [128]

Answer:

£165

Step-by-step explanation:

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3 years ago

Read 2 more answers

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Tomtit [17]

a) The cost for 10 mile taxi ride is $18

b) The cost for m mile taxi ride is 1.5 m + 3

Step-by-step explanation:

Step 1 :

The fixed charges for the pick up = $3

Charges per mile = $1.50

Let s denote the total miles driven and t be the total cost for the trip

This can be represented by the equation

t = 3 + 1.5s

Step 2:

Distance traveled by Jonathan in his trip = 10 miles

So cost for riding 10 miles is

t = 3 + 1.5(10) = 3 + 15 = $18

The cost for 10 mile taxi ride is $18

Step 3 :

If the distance traveled is m miles, then substituting s = m in the above equation we get the cost as 1.5 m + 3

Step 4 :

Answer :

a) The cost for 10 mile taxi ride is $18

b) The cost for m mile taxi ride is 1.5 m + 3

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3 years ago

What is equivalent <br> to 2x-7(6)

trasher [3.6K]

-40 i think.............. im not positive but i did my best

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3 years ago

Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0

Debora [2.8K]

So, the definite integral $\int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74$

Given that

$\int\limits^1_0 {x^{2} } \, dx = 13$

We find

$\int\limits^1_0 {(4 - 6x^{2} )} \, dx$

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, $Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx$

So, $\int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6\int\limits^1_0 {x^{2} } \, dx$