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lianna [129]
3 years ago
10

B + 4 = 2b - 5 B= ??

Mathematics
2 answers:
Sergio039 [100]3 years ago
6 0

Answer:

B=9

Step-by-step explanation:

b+4=2b-5

b-2b+4=-5

b-2b=-5-4

-b=-5-4

-b=-9

b=9

Andrej [43]3 years ago
3 0
The answer is B=-3.
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Which expression represents the composition [g o f o h](x) for the functions below?
spayn [35]

Answer:

5(15x-4)^3

Step-by-step explanation:

f(x) = 5x - 4

g(x) = 5x^3

h(x) = 3x

(gofoh)(x)=g(f(h(x)))

Replace h(x) first

g(f(h(x)))=g(f(3x))

Now we plug in 3x in f(x)

g(f(h(x)))=g(f(3x))=g(5(3x)-4)=g(15x-4)

Now we plug in 15x-4 in g(x)

g(f(h(x)))=g(f(3x))=g(5(3x)-4)=g(15x-4)=5(15x-4)^3

7 0
3 years ago
Add:(3x2 - 5x + 6) + (9 - 8x - 4x2)​
finlep [7]

You can see by the image how did I get the answer

3 0
4 years ago
(a) Use the reduction formula to show that integral from 0 to pi/2 of sin(x)^ndx is (n-1)/n * integral from 0 to pi/2 of sin(x)^
Sedbober [7]
Hello,

a)
I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\

= [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\

=0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\

= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\
= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx\\


I(1+n-1)= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\
I= \dfrac{n-1}{n} *\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\


b)
\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\
= \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\
= \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3} \\






\int\limits^{ \frac{\pi}{2} }_0 {sin^{5}(x)} \, dx \\
= \dfrac{4}{5}*\dfrac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx = \dfrac{8}{15}\\







c)

I_n=  \dfrac{n-1}{n} * I_{n-2} \\

I_{2n+1}=  \dfrac{2n+1-1}{2n+1} * I_{2n+1-2} \\
= \dfrac{2n}{2n+1} * I_{2n-1} \\
= \dfrac{(2n)*(2n-2)}{(2n+1)(2n-1)} * I_{2n-3} \\
= \dfrac{(2n)*(2n-2)*...*2}{(2n+1)(2n-1)*...*3} * I_{1} \\\\

I_1=1\\






3 0
4 years ago
3x-y=5 write the linear equation in slope-intercept form given
maxonik [38]

Answer:

y = -3x - 5

Step-by-step explanation:

3x - y = 5

+3x      +3x

-1y = 3x + 5

----  -----   -----

-1       -1     -1

y = -3x - 5


6 0
4 years ago
Find the area of the figures, HELP MEEE<br><br> no links, they cause bugs and viruses :3
Anna71 [15]

Answer:

93.6

Semi circle - 31.81 (rounded)

Step-by-step explanation:

find the area of the triangle which is 1/2 6*5.2 -=15.6

there are 6 triangles so that means it has to be multiplied by 6 - 15.6*6 =  93.6

Use the semi cirlce formula- (pi*r^2)/2

In this situation that is (pi*4.5^2 )/2 = 31.81 (rounded)

6 0
3 years ago
Read 2 more answers
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