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3 years ago

Explain how you can find 4 x 754 using two different methods

Mathematics

2 answers:

Citrus2011 [14]3 years ago

6 0

Solution: The value of $4\times 754$ is 3,016.

Explanation:

The first method of multiplication is given below:

Multiply the ones digit by the multiplier and write the ones digit of the result in the bottom line and add the tens digit to the next resultant, and same happens with the next number multiplied. This method show in given figure.

The second method of multiplication is given below:

Write the digits as the addition of their place values and then multiply those place value by multiplier individually after that add the results.

$4\times 754=4\times (700+50+4)\\4\times 754=4\times700+4\times50+4\times4\\4\times754=2800+200+16\\4\times754=3016$

Therefore, the value of $4\times 754$ is 3,016.

tresset_1 [31]3 years ago

5 0

The first method you can use is mental

754 * 4 = 3,016

The second method you can use is ...
21
754
4
*_____
3,016

4 * 4 = 16
5 * 4 = 20
7 * 4 = 28

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The spread of a virus is modeled by V (t) = −t 3 + t 2 + 12t,

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Functions can be used to model real life scenarios

The reasonable domain is $\mathbf{[0,\infty)}$ .
The average rate of change from t = 0 to 2 is 20 persons per week
The instantaneous rate of change is $\mathbf{V'(t) = -3t^2 + 2t + 12}$ .
The slope of the tangent line at point (2,V(20) is 10
The rate of infection at the maximum point is 8.79 people per week

The function is given as:

$\mathbf{V(t) = -t^3 + t^2 + 12t}$

(a) Sketch V(t)

See attachment for the graph of $\mathbf{V(t) = -t^3 + t^2 + 12t}$

(b) The reasonable domain

t represents the number of weeks.

This means that: t cannot be negative.

So, the reasonable domain is: $\mathbf{[0,\infty)}$

(c) Average rate of change from t = 0 to 2

This is calculated as:

$\mathbf{m = \frac{V(a) - V(b)}{a - b}}$

So, we have:

$\mathbf{m = \frac{V(2) - V(0)}{2 - 0}}$

$\mathbf{m = \frac{V(2) - V(0)}{2}}$

Calculate V(2) and V(0)

$\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}$

$\mathbf{V(0) = (0)^3 + (0)^2 + 12 \times 0 = 0}$

So, we have:

$\mathbf{m = \frac{20 - 0}{2}}$

$\mathbf{m = \frac{20}{2}}$

$\mathbf{m = 10}$

Hence, the average rate of change from t = 0 to 2 is 20

(d) The instantaneous rate of change using limits

$\mathbf{V(t) = -t^3 + t^2 + 12t}$

The instantaneous rate of change is calculated as:

$\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}$

So, we have:

$\mathbf{V(t + h) = (-(t + h))^3 + (t + h)^2 + 12(t + h)}$

$\mathbf{V(t + h) = (-t - h)^3 + (t + h)^2 + 12(t + h)}$

Expand

$\mathbf{V(t + h) = (-t)^3 +3(-t)^2(-h) +3(-t)(-h)^2 + (-h)^3 + t^2 + 2th+ h^2 + 12t + 12h}$ $\mathbf{V(t + h) = -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h}$

Subtract V(t) from both sides

$\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h - V(t)}$

Substitute $\mathbf{V(t) = -t^3 + t^2 + 12t}$

$\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h +t^3 - t^2 - 12t}$

Cancel out common terms

$\mathbf{V(t + h) - V(t)= -3t^2h -3th^2 - h^3 + 2th+ h^2 + 12h}$

$\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}$ becomes

$\mathbf{V'(t) = \lim_{h \to \infty} \frac{ -3t^2h -3th^2 - h^3 + 2th+ h^2 + 12h}{h}}$

$\mathbf{V'(t) = \lim_{h \to \infty} -3t^2 -3th - h^2 + 2t+ h + 12}$

Limit h to 0

$\mathbf{V'(t) = -3t^2 -3t\times 0 - 0^2 + 2t+ 0 + 12}$

$\mathbf{V'(t) = -3t^2 + 2t + 12}$

(e) V(2) and V'(2)

Substitute 2 for t in V(t) and V'(t)

So, we have:

$\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}$

$\mathbf{V'(2) = -3 \times 2^2 + 2 \times 2 + 12 = 4}$

Interpretation

V(2) means that, 20 people were infected after 2 weeks of the virus spread

V'(2) means that, the rate of infection of the virus after 2 weeks is 4 people per week

(f) Sketch the tangent line at (2,V(2))

See attachment for the tangent line

The slope of this line is:

$\mathbf{m = \frac{V(2)}{2}}$

$\mathbf{m = \frac{20}{2}}$

$\mathbf{m = 10}$

The slope of the tangent line is 10

(g) Estimate V(2.1)

The value of 2.1 is

$\mathbf{V(2.1) = (-2.1)^3 + (2.1)^2 + 12 \times 2.1}$

$\mathbf{V(2.1) = 20.35}$

(h) The maximum number of people infected at the same time

Using the graph, the maximum point on the graph is:

$\mathbf{(t,V(t) = (2.361,20.745)}$

This means that:

The maximum number of people infected at the same time is approximately 21.

The rate of infection at this point is:

$\mathbf{m = \frac{V(t)}{t}}$

$\mathbf{m = \frac{20.745}{2.361}}$

$\mathbf{m = 8.79}$

The rate of infection is 8.79 people per week

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