Ask question

Ask question

All categories

3 years ago

15

A classmate believes that a triangle can have both a right angle and an obtuse angle. Which of the following statements would yo

u pick to convince him that he is incorrect
(a) It is not possible because the sum of the angles in the triangle can’t exceed 180°
(b)It is not possible because if a triangle has an obtuse triangle, that means all the angles are obtuse angles. (c) It is not possible because if a triangle has a right triangle, that means all the angles are right angles (d)It is possible. The classmate is correct

1 answer:

Fofino [41]3 years ago

6 0

(a) It is not possible because the sum of the angles in the triangle can’t exceed 180°

You might be interested in

Based on the table of values below, find the slope between points where x = 5 and

aleksklad [387]

Answer:

Slope = 1

Step-by-step explanation:

The table gives the values of y corresponding to the values of x.

At x =5, y = 2.

At x = 7, y = 6 and

At x = 10, y = 7

Now, we have to determine the slope between points where x = 5 and where x = 10.

Now, the two concerned points are (5,2) and (10,7) and we have to find the slope between those two points.

Now, slope, $m = \frac{7 - 2}{10 - 5} = 1$ (Answer)

3 0

2 years ago

Read 2 more answers

What's 2x1 1/4x1 1/2 <br> I need this answer

kolezko [41]

The answer would be 3.75

8 0

2 years ago

Read 2 more answers

Find the exact circumference of a circle with the given radius. 5 inches C = 2.5π in. 10π in. 25π in.

docker41 [41]

The circumference, C, of a given circle may be calculated through the equation,

C = 2πr

where r is the radius.

Substituting the known values from the given in this item,
C = 2(π)(5 in)
<em>C = 10π in</em>

Thus, the answer is this item is the second choice.

5 0

2 years ago

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago

Expand 3(t -2) 3(t-2)=

iogann1982 [59]

Answer:

I hope this helps:)

Step-by-step explanation:

3 0

2 years ago