All categories

2 years ago

What is the first error Holden made in his proof?

Mathematics

2 answers:

kenny6666 [7]2 years ago

8 0

Answer:

Holden wanted to build “Australia's Own Car” and in 1948 the first Holden car was produced.

Hope this helps, pls mark me as brainliest

Dafna11 [192]2 years ago

3 0

Answer:

Step-by-step explanation:

He didn’t show his work.

You might be interested in

Please help. I am having trouble with this question. This is sixth-grade mathematics and is worth 50 points. I will mark brainli

antiseptic1488 [7]

Answer:

x is greater than or equal to 3

8 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

What is the answer to this.. 8(4n+13)=6n ...i need help

ollegr [7]

8(4n + 13) = 6n
Use distributive property so :8 (4n) = 32n and 8(13) = 104
32n + 104 = 6n
Subtract 32n from each side and you will get:
104 = 26n
then you divide each side by 26 and your answer will be:
n = 4

3 0

3 years ago

Read 2 more answers

A recipe for cookies uses sugar and baking powder. The recipe uses uses 3/4 cups of sugar and 1/2 teaspoon of baking powder

satela [25.4K]

Answer:

5/4 or 1 . 1/4

Step-by-step explanation:

3/4 + 1/2 = 5/4 or 1 . 1/4

4 0

3 years ago

7-41= (-15)+33= 62-84= (-26)-14=

Gala2k [10]

Answer:

-34

-22

-40

Explanation:

8 0

3 years ago