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3 years ago

12

8% of the High School students have their license. If there are 1000 high school students in your neighborhood, how many have th

eir license internet Enter your answer in the sface below. ??? teachers

1 answer:

ivolga24 [154]3 years ago

3 0

Answer:

80.

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Solve for j.<br> 4 - 2 = 3<br> = 3(1/3j+2)

Lerok [7]

J would equal 8/3 hope this helps

8 0

2 years ago

4/6 in the lowest term helpppppppppp!!!!!!!

Agata [3.3K]

2/3 would be the answer hope this helps.

3 0

3 years ago

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Estimate the integral ∫6,0 x^2dx by the midpoint estimate, n = 6

Anettt [7]

Splitting up the interval [0, 6] into 6 subintervals means we have

$[0,1]\cup[1,2]\cup[2,3]\cup\cdots\cup[5,6]$

and the respective midpoints are $\dfrac12,\dfrac32,\dfrac52,\ldots,\dfrac{11}2$ . We can write these sequentially as ${x_i}^*=\dfrac{2i+1}2$ where $0\le i\le5$ .

So the integral is approximately

$\displaystyle\int_0^6x^2\,\mathrm dx\approx\sum_{i=0}^5({x_i}^*)^2\Delta x_i=\frac{6-0}6\sum_{i=0}^5({x_i}^*)^2=\sum_{i=0}^5\left(\frac{2i+1}2\right)^2$

Recall that

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$
$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$
$\displaystyle\sum_{i=1}^n1=n$

so our sum becomes

$\displaystyle\sum_{i=0}^5\left(\frac{2i+1}2\right)^2=\sum_{i=0}^5\left(i^2+i+\frac14\right)$
$=\displaystyle\frac{5(6)(11)}6+\frac{5(6)}2+\frac54=\frac{143}2$

8 0

2 years ago

Need help with inscribed angles If M

Bogdan [553]

Given:

In the given circle O, BC is diameter, OA is radius, DC is a chord parallel to chord BA and $m\angle BCD=30^\circ$ .

To find:

The $m\angle AOB$ .

Solution:

If a transversal line intersect two parallel lines, then the alternate interior angles are congruent.

We have, DC is parallel to BA and BC is the transversal line.

$\angle OBA\cong \angle BCD$ [Alternate interior angles]

$m\angle OBA=m\angle BCD$

$m\angle OBA=30^\circ$

In triangle AOB, OA and OB are radii of the circle O. It means OA=OB and triangle AOB is an isosceles triangle.

We know that base angles of an isosceles triangle are congruent.

$\angle OAB\cong \angle OBA$ [Base angles of an isosceles triangle]

$m\angle OAB=m\angle OBA$

$m\angle OAB=30^\circ$

In triangle AOB,

$m\angle OAB+m\angle OBA+m\angle AOB=180^\circ$

$30^\circ+30^\circ+m\angle AOB=180^\circ$

$60^\circ+m\angle AOB=180^\circ$

$m\angle AOB=180^\circ-60^\circ$

$m\angle AOB=120^\circ$

Therefore, the measure of angle AOB is 120 degrees.

6 0

2 years ago

Just 6 pleaseeeeeeeeeee

Lynna [10]

Answer:

x>10

Step-by-step explanation:

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3 years ago

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