I need help with these 2 problems

Of the sixth graders in a school, 4/5 play at least one sport. Of those 2/3 play on a team. What fraction of 6th graders play a

artcher [175]

8/15 of 6th graders! 2/3 x 4/5 = 8/15

7 0

4 years ago

25 Pts + Brainliest for the first correct answer.

GuDViN [60]

Answer:

29) x = 17

30) x =14

31) x = 23

32) x = 44

Step-by-step explanation:

29)

6x - 17 = 85

6x = 102

x = 17

30)

4x + 21 + 103 = 180

4x + 124 = 180

4x = 56

x = 14

31)

2x + 23 = 4x - 23

-2x = - 46

x = 23

32)

4x - 42 = 134

4x = 176

x = 44

6 0

3 years ago

Please help me i don’t understand this question at all.

Maru [420]

9514 1404 393

Answer:

D.

Step-by-step explanation:

The wording "when x is an appropriate value" is irrelevant to this question. That phrase should be ignored. (You may want to report this to your teacher.)

When you look at the answer choices, you see that all of them are negative except the last one (D). When you look at the problem fraction, you see that it is positive.

The only reasonable choice is D.

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Your calculator can check this for you.

√12/(√3 +3) ≈ 3.4641/(1.7321 +3)

= 3.4641/4.7321 ≈ 0.7321 = -1 +√3

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If you want to "rationalize the denominator", then multiply numerator and denominator by the conjugate of the denominator. The conjugate is formed by switching the sign between terms.

$\displaystyle\frac{\sqrt{12}}{\sqrt{3}+3}=\frac{\sqrt{12}}{(\sqrt{3}+3)}\cdot\frac{(\sqrt{3}-3)}{(\sqrt{3}-3)}=\frac{\sqrt{36}-3\sqrt{12}}{3-9}\\\\=\frac{6-3\cdot2\sqrt{3}}{-6}=\boxed{-1+\sqrt{3}}$

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<em>Additional comment</em>

We "rationalize the denominator" in this way to take advantage of the relation ...

(a -b)(a +b) = a² -b²

Using this gets rid of the irrational root in the denominator, hence "rationalizes" the denominator.

We could also have multiplied by (3 -√3)/(3 -√3). This would have made the denominator positive, instead of negative. However, I chose to use (√3 -3) so you could see that all we did was change the sign from (√3 +3).

3 0

3 years ago

Suppose that

juin [17]

2000 dollars is owed with 9% interest rate.
Now, assuming that for 10 years, it is not being paid.
Let's calculate for the money that is now owed.
=> 2000 dollars * . 09 = 180 dollars for 1 years
=> 180 dollars * 10 years = 1800 dollars
=> 2000 + 1800 = 3800 dollars for 10 years.

6 0

3 years ago

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$

so, $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

since $\sqrt{3}=1.732$

so, $(6+7(1.732),0)\text{ and }(6-7(1.732),0)$ are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

3 0

3 years ago