Solve the following for y: y-2=-6(x-(-2))

What is the value of 13.997x4 to 2 decimal places ?

VARVARA [1.3K]

Answer:

Hence the answer to 2 decimal places becomes 55.99

Step-by-step explanation:

SOLUTION

We want to solve:

13.997×4 to 2 decimal places

Now:

13.997×4=55.988

Now, 55.988 to decimal places, after the decimal point, we count the first two numbers, which are 9 and then 8. Then check the number after the 8, is it up to 5? Yes, it is, because it is 8. Now change the 8 to 1 and add it to that second number after the decimal point, this gives nine.

Hence the answer to 2 decimal places becomes 55.99

5 0

2 years ago

A total of 133 million square miles of the Earth's surface are covered in water. The ratio of the area covered in water to the a

Anni [7]

Answer:

57 million square miles

Step-by-step explanation:

Let the area of water is w and area of land is l.

Their ratio is:

w/l = 7/3

or

133 / l = 7/3

Find the value of l

l = 133*3/7
l = 57 million square miles

6 0

2 years ago

The dimensions of a patio are shown above what is the area in square feet of the patio?

Verizon [17]

Answer:

uhh the picture is sideways .

5 0

3 years ago

Evaluate the triple integral ∭ExydV where E is the solid tetrahedon with vertices (0,0,0),(5,0,0),(0,9,0),(0,0,4).

Elan Coil [88]

Answer: $\int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV$ = 1087.5

Step-by-step explanation: To evaluate the triple integral, first an equation of a plane is needed, since the tetrahedon is a geometric form that occupies a 3 dimensional plane. The region of the integral is in the attachment.

An equation of a plane is found with a point and a normal vector. Normal vector is a perpendicular vector on the plane.

Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = $\left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]$ $\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]$

n = 36i + 0j + 45k - (0k + 0i - 20j)

n = 36i + 20j + 45k

Equation of a plane is generally given by:

$a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0$

Then, replacing with point P and normal vector n:

$36(x-5) + 20(y-0) + 45(z-0) = 0$

The equation is: 36x + 20y + 45z - 180 = 0

Second, in evaluating the triple integral, set limits:

In terms of z:

$z = \frac{180-36x-20y}{45}$

When z = 0:

$y = 9 + \frac{-9x}{5}$

When z=0 and y=0:

x = 5

Then, triple integral is:

$\int\limits^5_0 {\int\limits {\int\ {xy} \, dz } \, dy } \, dx$

Calculating:

$\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx$

$\int\limits^5_0 {\int\limits {\int\ {xy(\frac{180-36x-20y}{45} - 0 )} \, dy } \, dx$

$\frac{1}{45} \int\limits^5_0 {\int\ {180xy-36x^{2}y-20xy^{2}} \, dy } \, dx$

$\frac{1}{45} \int\limits^5_0 {90xy^{2}-18x^{2}y^{2}-\frac{20}{3} xy^{3} } \, dx$

$\frac{1}{45} \int\limits^5_0 {2430x-1458x^{2}+\frac{94770}{125} x^{3}-\frac{23490}{375}x^{4} } \, dx$

$\frac{1}{45} [30375-60750+118462.5-39150]$

$\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx$ = 1087.5

The volume of the tetrahedon is 1087.5 cubic units.

3 0

3 years ago

Solve the unknown angle measure? Help Any1 ASAP! Will Mark Brainliest!

zlopas [31]

Answer:

X=40°

X=30°

X=50°

Step-by-step explanation:

Let our unknown angles be denoted by $X$

Part I

We are given the sum of the angles as 70°, the known as 30° and the unknown as X;

To find X, we subtract the known angle from the sum as:

X=70°-30°=40°

Hence X= 40°

Part II

We are given the sum of the angles as 70°, the known as 40° and the unknown as X;

To find X, we subtract the known angle from the sum as:

X=70°-40°=30°

Hence X= 30°

Part III

We are given the sum of the angles as 80°, the known as 30° and the unknown as X;

To find X, we subtract the known angle from the sum as:

X=80°-30°=50°

Hence X= 50°

3 0

3 years ago