Henry draws an obtuse triangle how many obtuse angles does Henry's triangle have?

7/2 ≥ b + 9/5. and btw, the ones that has the slashes are fractions. :3

Ainat [17]

Answer:

I believe it's

* B. b<1 7/10

I'm sorry if its wrong

6 0

3 years ago

Evaluate the indefinite integral. (Use C for the constant of integration.) sin(10x) 1 cos2(10x) dx

Schach [20]

Answer:

The result of the integral is $-\frac{\cos^3{10x}}{30} + C$

Step-by-step explanation:

We are given the following integral:

$\int \sin{(10x)}\cos^{2}{(10x)} dx$

I am going to solve by substitution, using $u = \cos{10x}$ , so $du = -10\sin{10x}dx, dx = -\frac{du}{10\sin{10x}}$ . So, we have

$-\frac{1}{10} \int u^2 du$

Which has the following result:

$-\frac{u^3}{30} + C$

Going back to x, the result of the integral is

$-\frac{\cos^3{10x}}{30} + C$

5 0

3 years ago

Last week, a coral reef grew 19.8mm taller. How much did it grow in meters?

Oliga [24]

Answer:

Last week the coral reef grew .0196m

Step-by-step explanation:

So it's hard to explain but simply Milli witch is the symbol "m" is a unit prefix in the metric system denoting a factor of one thousandth (10 −3). So 19.6mm is 19.6*10^-3m. in other words 19.6 times 10^-3m

hope this helps sorry it's so complicated :/

5 0

3 years ago

Read 2 more answers

Can someone solve what y=?

ladessa [460]

Answer:

y = 1

Step-by-step explanation:

This is because the line intersects the y-axis at 1.

4 0

3 years ago

Read 2 more answers

Add or subtract. Write the result in the form a+bi

STALIN [3.7K]

Number 1: (-1+2i)+(6-9i)

$Group\:the\:real\:part\:and\:the\:imaginary\:part\:of\:the\:complex\:number\\\left(a+bi\right)\pm \left(c+di\right)=\left(a\:\pm \:c\right)+\left(b\:\pm \:d\right)i\\=\left(-1+6\right)+\left(2-9\right)i\\=5-7i$

Number 2: (3-3i)-(4+7i)

$Group\:the\:real\:part\:and\:the\:imaginary\:part\:of\:the\:complex\:number\\\left(a+bi\right)\pm \left(c+di\right)=\left(a\:\pm \:c\right)+\left(b\:\pm \:d\right)i\\=\left(3-4\right)+\left(-3-7\right)i\\Refine\\=-1-10i$

Number 3: (-5+2i)+(-2+8i)

$Group\:the\:real\:part\:and\:the\:imaginary\:part\:of\:the\:complex\:number\\\left(a+bi\right)\pm \left(c+di\right)=\left(a\:\pm \:c\right)+\left(b\:\pm \:d\right)i\\=\left(-5-2\right)+\left(2+8\right)i\\Refine\\=-7+10i$

8 0

3 years ago