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kap26 [50]
3 years ago
9

Whats nine рlus ten ?

Mathematics
1 answer:
EleoNora [17]3 years ago
7 0

Answer:

19

Step-by-step explanation:

9+10=19

You might be interested in
What is 3x+4y=16 for x
sergiy2304 [10]
3x + 4y = 16           Write original equation     
3x + 4y - 4y = -4y + 16          Subtract 4y from each side
3x = -4y +16                    Simplify
3x/3 = -4y/3 + 16/3      Divide each side by three
x = -4y/3 +16/3             Simplify

I hope this helps!

4 0
3 years ago
10th term of an= 20-2n
Dmitry [639]

Answer:

We have that a_{n} = 20-2n

Step-by-step explanation:

We have just to replace n=10 on the general equation

a_{10} = 20-2×10

Therefore:

a_{10} = 0

8 0
3 years ago
Find function inverse of f? F(x)=x+2/7
nignag [31]

Answer:

f’(x) = x -2/7

Step-by-step explanation:

Let f(x)= y

Thus;

y = x + 2/7

x = y-2/7

So the inverse f’(x) = x - 2/7

6 0
3 years ago
What’s the area of this figure
kvasek [131]

Answer:

743.25m^2

Step-by-step explanation:

Area of triangle = 1/2bh

= 1/2 x 30 x 26

= 390

Area of semi circle = 1/2 πr^2

= 1/ 2 x 3.14 x 15 x15

=353.25m^2

= 390 + 353.25

= 743.25m^2

8 0
3 years ago
Read 2 more answers
Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)
USPshnik [31]
See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) +  f(\frac{x_{1} + x_{2} }{2}) + ... +  f(\frac{x_{n-1} + x_{n} }{2})]

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
M_{4} = \frac{1 - 0}{4} [ f(\frac{0 +  \frac{1}{4} }{2}) +  f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) +  f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]
M_{4} = \frac{1}{4} [ f(\frac{1}{8}) +  f(\frac{3}{8}) +  f(\frac{5}{8}) + f(\frac{7}{8})]

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]

M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]

M₄ = 1/4 · (2 - 478/512)
     = 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0
4 years ago
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