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Nady [450]
3 years ago
10

frac r9+8−13= 9r​ +8" alt="-13 =\dfrac r9+8−13= 9r​ +8" align="absmiddle" class="latex-formula">

Mathematics
2 answers:
Murljashka [212]3 years ago
7 0
Don’t know but it I’ll be here with a big boi 2
just olya [345]3 years ago
4 0

Answer:

r = -189

Step-by-step explanation:

multiply both sides by 9 to get rid of the fraction.

-117=r+72

now subtract 72 from both sides.

-117-72=r

r = -189

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Help! What is the area of each shape?
choli [55]

Answer:

27.84 m² , 42.48 m²

Step-by-step explanation:

The area (A) of a triangle is calculated as

A = \frac{1}{2} bh ( b is the base and h the perpendicular height )

Shape A

has b = 9.6 and h = 5.8, then

A = 0.5 × 9.6 × 5.8 = 27.84 m²

Shape B

has b = 11.8 and h = 7.2 , then

A = 0.5 × 11.8 × 7.2 = 42.48 m²

7 0
2 years ago
Read 2 more answers
Find the value of x. Round to the nearest tenth. tan 46 = x/12
otez555 [7]
Tan46=x/12
12×tan 46=x
12×1.03=x
0.21255=x

5 0
3 years ago
Read 2 more answers
Please help me!! thank you!!
I am Lyosha [343]

Answer:

A. 106 + 2x + 50 = 180

Step-by-step explanation:

A. 106 + 2x + 50 = 180. Correct.

B. 2x + 50 = 180. Incorrect. Angle is not a Straight Angle.

C. 2x + 50 = 106. Incorrect. That equals the first angle.

hope this helps.

4 0
2 years ago
Use the Fundamental Theorem for Line Integrals to find Z C y cos(xy)dx + (x cos(xy) − zeyz)dy − yeyzdz, where C is the curve giv
Harrizon [31]

Answer:

The Line integral is π/2.

Step-by-step explanation:

We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:

f_x = ycos(xy)

f_y = xcos(xy) - ze^{yz}

f_z = -ye^{yz}

we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence

\int{ycos(xy)} \, dx = \int cos(u) \, du = sen(u) + C = sen(xy) + C(y,z)  

(Remember that c is treated like a constant just for the x-variable).

This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.

\int {-ye^{yz}} \, dy = \int {-e^{u} \, dy} = -e^u +K = -e^{yz} + K(z)

Where, again, the constant of integration depends on Z.

As a result,

f(x,y,z) = cos(xy) - e^{yz} + K(z)

if we derivate f over z, we obtain

f_z(x,y,z) = -ye^{yz} + d/dz K(z)

That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)

The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.

3 0
3 years ago
How do I do this please help!
jeka94

Answer:

put the 4x on the top and the negative 3 on the left not the bottom

6 0
3 years ago
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