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3 years ago

What is the value of h in the triangle below

Mathematics

1 answer:

Nesterboy [21]3 years ago

4 0

Answer:

Step-by-step explanation:

8 or 6

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Solve this right now for a billion dollars

Tema [17]

Answer:

Step-by-step explanation:

Simplifying

2.1x + -5.7 = 3.9(4.3 + -1.8x)

Reorder the terms:

-5.7 + 2.1x = 3.9(4.3 + -1.8x)

-5.7 + 2.1x = (4.3 * 3.9 + -1.8x * 3.9)

-5.7 + 2.1x = (16.77 + -7.02x)

Solving

-5.7 + 2.1x = 16.77 + -7.02x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '7.02x' to each side of the equation.

-5.7 + 2.1x + 7.02x = 16.77 + -7.02x + 7.02x

Combine like terms: 2.1x + 7.02x = 9.12x

-5.7 + 9.12x = 16.77 + -7.02x + 7.02x

Combine like terms: -7.02x + 7.02x = 0.00

-5.7 + 9.12x = 16.77 + 0.00

-5.7 + 9.12x = 16.77

Add '5.7' to each side of the equation.

-5.7 + 5.7 + 9.12x = 16.77 + 5.7

Combine like terms: -5.7 + 5.7 = 0.0

0.0 + 9.12x = 16.77 + 5.7

9.12x = 16.77 + 5.7

Combine like terms: 16.77 + 5.7 = 22.47

9.12x = 22.47

Divide each side by '9.12'.

x = 2.463815789

Simplifying

x = 2.463815789

5 0

3 years ago

Read 2 more answers

Evaluate the integral. W (x2 y2) dx dy dz; W is the pyramid with top vertex at (0, 0, 1) and base vertices at (0, 0, 0), (1, 0,

In-s [12.5K]

Answer:

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

Step-by-step explanation:

Given that:

$\iiint_W (x^2+y^2) \ dx \ dy \ dz$

where;

the top vertex = (0,0,1) and the base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 \int ^{1-z}_0 (x^2+y^2) \ dx \ dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 ( \dfrac{(1-z)^3}{3}+ (1-z)y^2) dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^3}{3} \ y + \dfrac {(1-z)y^3)}{3}] ^{1-x}_{0}$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^4}{3}+ \dfrac{(1-z)^4}{3}) \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =\dfrac{2}{3} \int^1_0 (1-z)^4 \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =- \dfrac{2}{15}(1-z)^5|^1_0$

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

7 0

3 years ago

Which option below best describes the maximums of these two functions? Function g has the greater maximum of 2. Functions g and

MissTica

Answer:

salak

Step-by-step explanation:

5 0

3 years ago

Help with number 9 please please thank you

Harlamova29_29 [7]

Answer:

$\large\boxed{53.9ml=0.0539l}$

Step-by-step explanation:

$1l=1,000ml\to1ml=\dfrac{1}{1,000}l=0.001l\\\\\text{Therefore}\\\\53.9ml=53.9\cdot0.001l=0.0539l$

6 0

3 years ago

A plane is flying at the height of 5000 meter above the sea level. at a particular point, it is excatly above a submarine floati

AnnyKZ [126]

Answer:

3800 meters

Step-by-step explanation:

3 0

3 years ago