Solve for b.

Cos15°°- Sin 15°= 1/√2

ololo11 [35]

<h2>Refer this attachment</h2>

Hope it's help you

4 0

3 years ago

HELP !!!! Equations with 1 no or infinite solutions

ryzh [129]

1. v=0

2. No solution

3. m=0

4. No solution

5. p= -7

6. x=5

7. -12 = - 12

8. x=5

9. 3=3

10. -32= -32

11. x=0

12. No solution

I'm sorry if anything is wrong.

6 0

2 years ago

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Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

8 0

3 years ago

I am thinking of two numbers. The difference of the two numbers is 36. Their sum is 286. What are the two numbers I am thinking

Diano4ka-milaya [45]

Let's call the two numbers a and b. So we know that the difference of the two numbers is 36:

a - b = 36

And we know that their sum is 286:

a + b = 286

We can use either of these equations to solve for one of the variables in terms of the other variable. Let's use the first and solve for a:

a - b = 36

a = 36 + b

Now we plug this into the other equation and solve for b there:

(36 + b) + b = 286

36 + 2b = 286

2b = 250

b = 125

6 0

3 years ago

Read 2 more answers

Can someone please help me I really need help please help me

scZoUnD [109]

Answer:

18.87 square cm.

Step-by-step explanation:

The area of the rectangle will be (4 + 4) * 4, since the length of the rectangle would be the diameter of the circle, and the width of the rectangle would be the radius. (4 + 4) * 4 = 8 * 4 = 32 square cm.

Then, we can calculate the area of the semicircle. The area of a circle is pi * r^2, so the area of a semicircle will be half of that. pi * (4^2) / 2 = pi * 16 / 2 = 8pi. 8 * 3.14159265 = 25.1327412 square cm.

The shaded area of the middle of the shape will then be 32 - 25.1327412 = 6.8672588 square cm.

The two triangles will have the same area. Their bases will be 14 minus the diameter of the circle, then divide that by 2 to get each separate base. 14 - 8 = 6 / 2 = 3. The heights of the triangles will be the radius of the circle, or 4 cm.

1/2 * 3 * 4 = 1/2 * 12 = 12/2 = 6. That is the area of one triangle, so the area of both triangles would be 6 * 2 = 12 square cm.

6.8672588 + 12 = 18.8672588, or 18.87 square cm.

Hope this helps!

3 0

2 years ago

Read 2 more answers