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Bogdan [553]
3 years ago
8

write about which method you prefer to use to divide by 5 counting up, counting back on a number line, or dividing by 10, and th

en doubling the quotient?
Mathematics
2 answers:
vazorg [7]3 years ago
5 0

I usually prefer to use the counting up, counting back on the number line because it is simpler and more efficient since it is like counting 1, 2, 3, but using 5 as interval.

Hope this answer will be a good help for you.

maria [59]3 years ago
4 0

Answer:

In First method : counting up, counting back on a number line,

If we want the quotient after dividing the number by 5 then we count how many 5 we get from 0 to the dividend.

For example :  \frac{30}{5}

Since, from 0 to 30 there are six 5's obtained. ( because 5 × 6 = 30 )

Thus,  \frac{30}{5}=6

In Second Method : dividing by 10, and then doubling the quotient.

First we divide the number by 10 then multiply the quotient by 2.

For Example: \frac{30}{5}

Since, \frac{30}{10}=3

2\times 3 = 6

Thus, \frac{30}{5}=6

Now, when we compare the above methods then we conclude that for the smaller numbers first method is appropriate because for small numbers we can easily count total 5's from 0. While for large numbers Second method is appropriate because it is hard to count the total 5's for the large number.


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Solve each pair of simultaneous equations by substitution method
WARRIOR [948]

Answer:

x = 1       y = -4

Step-by-step explanation:

- Plug the value of y into the other equation.

x + 3y = -11

x + 3(-4x) = -11

x - 12x = -11

-11x = -11

x = 1

- Now substitute the value of x into any equation.

y = -4x

y = -4(1)

y = -4

3 0
3 years ago
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7m+n over q= 2m. How do I solve for q
ryzh [129]
I believe you have to get all the like terms on the same side. So since 7m+n is being divided by Q in order to move it to the other side of the equal sign you must multiply each side by Q. Then you would end up with 7m+n=2m(q). 7m and 2m are alike so you then take the 2m and subtract it from each side. So your equation should look like 7m-2m+n=Q. Then combine your like terms and you end up with 5m+n=q 
7 0
3 years ago
The polynomial equation x^4 + x^3 + x^2 + x + 1 = 0 has zero positive real roots. Please select the best answer from the choices
const2013 [10]

There are zero positive real roots for the given polynomial equation x^4 + x^3 + x^2 + x + 1 = 0. This is explained by Descarte's rule of signs. So, the best choice is T (true).

<h3>What is Descarte's rule of signs?</h3>
  • Descarte's rule of signs tells about the number of positive real roots and negative real roots.
  • The number of changes in signs of the coefficients of the terms of the given polynomial f(x) gives the positive real zeros of the polynomial.
  • The number of changes in signs of the coefficients of the terms of the given polynomial when f(-x) gives the negative real zeros of the polynomial.

<h3>Calculation:</h3>

The given polynomial equation is x^4 + x^3 + x^2 + x + 1 = 0

On applying Descarte's rule of signs,

f(x)=x^4 + x^3 + x^2 + x + 1

Since there are no changes in the signs of the coefficients of any of the terms in the above polynomial, the polynomial has no positive real roots.

f(-x)=(-x)^4+(-x)^3+(-x)^2+(-x)+1\\      = x^4-x^3+x^2-x+1

Since there are four changes in the signs of the coefficients of the terms of the given polynomial when f(-x), the polynomial has 4 negative real roots.

Therefore, the given polynomial equation has zero positive real roots. So, the correct choice is T(true).

Learn more about Descarte's rule of signs here:

brainly.com/question/11590228

#SPJ1

5 0
2 years ago
F(4) if f(x) = 5x - 4
Aleks04 [339]

Answer:

16

Step-by-step explanation:

Plug in 4 for x

f(4) = 5(4) - 4 = 16

4 0
3 years ago
The data below are the temperatures on randomly chosen days during the summer and the number of employee absences at a local com
s2008m [1.1K]

Answer:

The critical region is t ≥ t(0.025, 7) = 2.365

Since the calculated value of t= 18.50249 falls in the critical region we reject the null hypothesis and conclude that there is sufficient reason to support the claim of a linear relationship between the two variables.

Step-by-step explanation:

We set up our hypotheses as

H0: β= 0   the two variable X and Y are not related

Ha: β  ≠ 0. the two variables X and Y are related.

The significance level is set at α =0.05

The test statistic if, H0 is true, is  t= b/s_b

Where   Sb =S_yx/√(∑(X-X`)^2 )

Syx = √((∑(Y-Y`)^2 )/(n-2))

In the given question we have the estimated regression line as y= 0.449x - 30.27

X Y X2         Y2      XY

72 3 5184 9    216

85 7 7225 49    595

91 10 8281 100      910

90 10 8100 100      900

88 8 7744 64       704

98 15 9604 225      1470

75 4 5625 16       300

100 15 10000 225       1500

<u>80 5 6400 25        400         </u>

<u>∑779 77 68163 813 6995</u>

Now finding the variances

∑(Y-Y`)^2  = ∑〖Y^2- a〗 ∑Y- b∑XY

                      = 813 – (- 30.27)77 - 0.449(6995)

                       = 813+2330.79 – 3140.755

                       = 3.035

∑(X-X`)^2 =  ∑X^2  – (∑〖X)〗^2 /n

                   = 68163 – (779)2/9

                    = 736.22

Syx = √((∑(Y-Y`)^2 )/(n-2))  = √(3.035/7) = 0.65846 and

Sb =S_yx/√(∑(X-X`)^2 ) = (0.65846  )/27.13337 = 0.024267

t= b/s_b  = 0.449/ 0.024267 = 18.50249

The critical region is t ≥ t(0.025, 7) = 2.365

Since the calculated value of t= 18.50249 falls in the critical region we reject the null hypothesis and conclude that there is sufficient reason to support the claim of a linear relationship between the two variables.

Download docx
6 0
3 years ago
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