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4 years ago

Please please help me

Mathematics

1 answer:

docker41 [41]4 years ago

7 0

Ur missing numbers are -2, 6, and 36

Since the center point is (-2,6), the numbers are reversed in the equation to: (x+2) or(x- -2) and (y-6)

The radius is 6 units so u square it to get 36

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Help plz:)))I’ll mark u Brainliest

Mamont248 [21]

Answer:

4/5

Step-by-step explanation:

Given :

A right angled triangle with sides 24 , 3 and 40 .

And we need to find the value of sinZ .

We know that , sine is the ratio of perpendicular and Hypotenuse. So that ,

$:\implies$ sinZ = p/h

$:\implies$ sin Z = 32/40

$:\implies$ sin Z = 4/5

<u>Hence the renquired answer is 4/5.</u>

7 0

3 years ago

Find th areas and circumference 5 cm 2 cm

mihalych1998 [28]

Answer:

The Area is 10cm, and the perimeter is 14cm

Step-by-step explanation:

Area:

a = w · l 5cm · 2cm =10cm

Perimeter:

p = 2(l+w) 2(5 + 2) = 14

7 0

2 years ago

Which equation has a graph that includes the point (4.5, 14)? Select all that apply. A. y = 2x + 5 B. y = 3x + 1.5 C. y = 4x – 4

erastovalidia [21]

Answer:

A and C

Step-by-step explanation:

Just plug in the point into the equations:

a) 14 = 2(4.5) + 5

14 = 9 + 5 14 = 14 A is correct

b) 14 = 3(4.5) + 1.5

14 = 13.5 + 1.5 14 ≠ 15 B is not correct

c) 14 = 4(4.5) - 4

14 = 18 - 4 14 = 14 C is correct

d) is not correct 12 is already to large and 10 is not a negative so it is far to large

3 0

3 years ago

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2