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Andreas93 [3]
3 years ago
11

I need this answered plz!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Fynjy0 [20]3 years ago
6 0

Answer:

Red: Left side equation

Blue: Right side equation

2x+7=2(x-1)

2x+7=2x-2

2x-2x=-2-7

0x=-9= 0 No solution

Step-by-step explanation:

Use Desmos graphing calculator if u need help with graphs

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Help<br><br><br><br><br><br> Thrhfjfhfhhfhfhfjf
12345 [234]

Answer: 5

Step-by-step explanation:

a^2 + b^2 = c^2

12^2 + x^2 = 13^2

144 + x^2 = 169

x^2 = 25

square root both sides

x = 5

to check, do 12^2 + 5^2 =169

sqrt of 169 is 13

hope this helps

4 0
3 years ago
Write the equation of the line, given the y- and x-intercepts:
sasho [114]

Answer:

y=2x+19

Step-by-step explanation:

Slope: 0-19/-9.5-0=-19/-9.5=2

y-intercept: 19

3 0
3 years ago
Read 2 more answers
Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16
4vir4ik [10]

Answer:

\mathbf{\iiint_E  E \sqrt{x^2+y^2} \ dV =\dfrac{81 \  \pi}{80}}

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

9 -16 (x^2 + y^2) = 0 \\ \\  16 (x^2 + y^2)  = 9 \\ \\  x^2+y^2 = \dfrac{9}{16}

x^2+y^2 = (\dfrac{3}{4})^2

where:

0 ≤ r ≤ \dfrac{3}{4} and 0 ≤ θ ≤ 2π

∴

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0}  \ \ \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2})  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0}  ( 9r^2-16r^4})  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0}  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0}  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}

\iiint_E  E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi

\iiint_E  E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi

\iiint_E  E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi

\mathbf{\iiint_E  E \sqrt{x^2+y^2} \ dV =\dfrac{81 \  \pi}{80}}

4 0
3 years ago
2. Solve the equation and check.
Andrej [43]
The answer is x=3 i believe
8 0
2 years ago
Raíz cuadrada de 110
lara31 [8.8K]

<span>La raíz cuadrada de 110 es aproximadamente 10.48809.
</span>
Espero que esto haya ayudado :)
6 0
3 years ago
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