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3 years ago

What is the STIMULUS in the following situation: All of the students go home or

Mathematics

1 answer:

kolbaska11 [484]3 years ago

7 0

I think it’s c the bell ringer

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How do you factorise x^2+2x-15

crimeas [40]

I hope this helps you

7 0

3 years ago

Will award brainliest to correct answer!

Artyom0805 [142]

Let's begin noting that a triangle is isosceles if and only if two of its angles are congruent. We can thus find the angle <ABP, recalling that the sum of the interior angles of a triangle is equal to 180°.

$\angle \: ABP = \frac{180 - 80}{2} = 50 \degree$

Finally, let point K be the intersection between segments BC and PQ, and let's note that the triangle PQB is a right isosceles triangle, since all the angles in a square are equal to 90°, and the two triangles APB and BQC are congruent.

Therefore, the angle BKQ is equal to 180-50-45=85°.

Of course angle BKP=180-85=95°.

Hope this helps :)

6 0

2 years ago

Bennett is setting up a light display for a wedding with two 10-foot strands of lights. The lights will be attached to the top o

Anuta_ua [19.1K]

Answer:

7.1

Step-by-step explanation:

write pythagorean theorem $x^{2}$ + $x^{2}$ = $10^2$

2 $x^2$ = 100

divide 2 both side:

$x^{2}$ = 50

$\sqrt{x^2}$ = $\sqrt{50}$

$x$ = $\sqrt{50}$

$x$ = 7.07106781

round to nearest tenth:

$x$ = 7.1

8 0

3 years ago

Bob recorded the number of hours each person watched TV in a week. He then organized the information in a frequency table.

seraphim [82]

Answer:

yque ha haceqr ailluey

Step-by-step explanation:

6 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago