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3 years ago

I was working on this question and and a bit confused, somebody plz help!! :)

Mathematics

2 answers:

ladessa [460]3 years ago

5 0

The answer should be 6s^3 because you add the two powers together since the number (6) is the same

pav-90 [236]3 years ago

3 0

Answer:

36s³

Step-by-step explanation:

Using the rule of exponents

$a^{m}$ × $a^{n}$ ⇔ $a^{(m+n)}$

Given

6s² × 6s

= 6 × s² × 6 × $s^{1}$

= 6 × 6 × s² × $s^{1}$

= 36 × $s^{(2+1)}$

= 36s³

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Use your device keyboard to enter x+1= 7 in the box.<br>Check Answer

egoroff_w [7]

<h2>ANSWER:</h2>

_______________________________________________________

Answer is x = 6

_______________________________________________________

<h2>EXPLANATION:</h2>

_______________________________________________________

We know that 7 - 1 is 6. 6 is what x equals because 7 - 1 is 6 and 6 + 1 is equal to 7.

Hope this helps!

7 0

3 years ago

Solve for X <br><br> -5/2 = -3 + 4/8x

allsm [11]

Answer:

Step-by-step explanation:

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3 years ago

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A disk with radius 3 units is inscribed in a regular hexagon. Find the approximate area of the inscribed disk using the regular

Oksi-84 [34.3K]

Answer:

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$

Step-by-step explanation:

we know that

we can divide the regular hexagon into 6 identical equilateral triangles

see the attached figure to better understand the problem

The approximate area of the circle is approximately the area of the six equilateral triangles

Remember that

In an equilateral triangle the interior measurement of each angle is 60 degrees

We take one triangle OAB, with O as the centre of the hexagon or circle, and AB as one side of the regular hexagon

Let

M ----> the mid-point of AB

OM ----> the perpendicular bisector of AB

x ----> the measure of angle AOM

$m\angle AOM =30^o$

In the right triangle OAM

$tan(30^o)=\frac{(a/2)}{r}=\frac{a}{2r}\\\\tan(30^o)=\frac{\sqrt{3}}{3}$

$\frac{a}{2r}=\frac{\sqrt{3}}{3}$

we have

$r=3\ units$

substitute

$\frac{a}{2(3)}=\frac{\sqrt{3}}{3}\\\\a=2\sqrt{3}\ units$

Find the area of six equilateral triangles

$A=6[\frac{1}{2}(r)(a)]$

simplify

$A=3(r)(a)$

we have

$r=3\ units\\a=2\sqrt{3}\ units$

substitute

$A=3(3)(2\sqrt{3})\\A=18\sqrt{3}\ units^2$

Therefore

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$

6 0

3 years ago

Nathan plants equal sized squares of sod in his front yard each square has an area of 6 square feet Nathan plants a total of 100

Darya [45]

Multiply 6x1000 wich will give you the total area

7 0

3 years ago

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Assume the competing hypotheses take the following form: H0: µ1 – µ2 = 0, HA: µ1 – µ2 ≠ 0, where µ1 is the population mean for p

DedPeter [7]

Answer:

$t=\frac{\bar x_1- \bar x_2}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } }$

Step-by-step explanation:

H0: µ1 – µ2 = 0

HA: µ1 – µ2 ≠ 0

We have given,

The population variances are not known and cannot be assumed equal.

The test statistic for the test is

$t=\frac{\bar x_1- \bar x_2}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } }$

Where,

$\bar x_1$ = sample meaan of population 1

$\bar x_2$ = sample mean of population 2

$n_1$ = sample size of population 1

$n_2$ = sample size of population 2

Therefore, this is the test

$t=\frac{\bar x_1- \bar x_2}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } }$

7 0

3 years ago