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3 years ago

What is the repeating or terminating decimal that is equivalent to 11/36

Mathematics

1 answer:

Oduvanchick [21]3 years ago

3 0

Answer:

0.30555555....

Step-by-step explanation:

(The fives keep on coming, this is a repeating decimal.)

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If 2(4x + 3)/(x - 3)(x + 7) = a/x - 3 + b/x + 7, find the values of a and b.

zmey [24]

Answer:

$a=3$ and $b=5$ .

Step-by-step explanation:

So I believe the problem is this:

$\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}$

where we are asked to find values for $a$ and $b$ such that the equation holds for any $x$ in the equation's domain.

So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).

In other words this will clear the fractions.

$\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)$

$2(4x+3)=a(x+7)+b(x-3)$

As you can see there was some cancellation.

I'm going to plug in -7 for x because x+7 becomes 0 then.

$2(4\cdot -7+3)=a(-7+7)+b(-7-3)$

$2(-28+3)=a(0)+b(-10)$

$2(-25)=0-10b$

$-50=-10b$

Divide both sides by -10:

$\frac{-50}{-10}=b$

$5=b$

Now we have:

$2(4x+3)=a(x+7)+b(x-3)$ with $b=5$

I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.

$2(4\cdot 3+3)=a(3+7)+b(3-3)$

$2(12+3)=a(10)+b(0)$

$2(15)=10a+0$

$30=10a$

Divide both sides by 10:

$\frac{30}{10}=a$

$3=a$

So $a=3$ and $b=5$ .

4 0

3 years ago

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Simplify the polynomial expressions by combining like terms and then multiply the resulting binomial expressions to find their p

patriot [66]

(6x-9)(8+5x-5)=(6x-9)(5x+3)=

$=6x\cdot5x-5x\cdot9+6x\cdot3-9\cdot3=\\=30x^2-27x-27$

6 0

3 years ago

Read 2 more answers

What is the value of x?<br> 16<br> 34

lord [1]

Answer:30

Step-by-step explanation:

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3 years ago

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Find two numbers if their sum is 115 and their difference is 7

kobusy [5.1K]

Answer:

54+61

Step-by-step explanation:

54 + 61 = 115

61 - 54 = 7

4 0

3 years ago

A teacher was interested in finding out whether a special study program would increase the scores of students on a national exam

bija089 [108]

Answer:

Check the explanation

Step-by-step explanation:

Going by the first attached image below we reject H_o against H_1 if obs. $T > t_{\alpha /2;n-1}$

here obs.T=1.879

$\therefore obs.T \ngtr 2.447=t_{0.025;6}$

we accept $H_o:\mu _{1}=\mu_{2}$ at 5% level of significance.

i.e there is no sufficient evidence to indicate that the special study program is more effective at 5% level of significance.

this problem is simillar to the previous one except the alternative hypothesis.

Let X_i's denote the bonuses given by female managers and Y_i's denote the bonuses given by male managers.

we assume that $X_i \sim N(\mu _{1},\sigma _{1}^{2}) Y_i \sim N(\mu _{2},\sigma _{2}^{2})$ independently

We want to test $H_0:\mu_{1}=\mu_{2} vs H_1:\mu_{1}\neq \mu_{2}$

define $D_i=X_i-Y_i , i=1(1)8$

now $D_i\sim N(\mu _{1}-\mu _{2}=\mu _{D},\sigma _{1}^{2}+\sigma _{2}^{2}=\sigma _{D}^{2}) , i=1(1)8$

the hypothesis becomes

$H_0:\mu_{D}=0 vs H_1:\mu_{D}\neq 0$

in the third attached image, we use the same test statistic as before

i.e at 5% level of significance there is not enough evidence to indicate a difference in average bonuses .

3 0

3 years ago