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jeka57 [31]
3 years ago
6

Find the area of the shape shown below.​

Mathematics
2 answers:
Alex787 [66]3 years ago
8 0

Answer:

7

Step-by-step explanation:

neonofarm [45]3 years ago
3 0

Answer:

7

Step-by-step explanation:

First ignore the cut corner and pretend that it is just a rectangle.

The rectangle's area would be 2×4=8.

Then find out the cut corner's area.

The area of that triangle would be 1.

8-1=7 So the area of the shape is 7.

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Nookie1986 [14]

Answer:

2.Goes in first box

4.goes in second  box

3.goes in the third box

1.goes in the fourth box

Step-by-step explanation:

3 0
3 years ago
What is 5/8+3/10 in simplest form?
NNADVOKAT [17]

Answer:

37/40

Step-by-step explanation:

5/8 +3/10

We need to get a common denominator of 40

5/8 *5/5 = 25/40

3/10 *4/4 = 12/40

25/40 + 12/40

37/40

4 0
3 years ago
Read 2 more answers
Combining like terms with negative coefficient -3x-6+(-1)
Andreyy89

Answer:

-3x-7

Step-by-step explanation:

So firs you would change the sign of +(-1)

-3x-6-1

-3x-7

5 0
3 years ago
Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

8 0
3 years ago
What is the scientific notation 3,400,000,000
Diano4ka-milaya [45]
3.4^9 is the answer/.............
6 0
4 years ago
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