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mel-nik [20]
3 years ago
13

Select the correct location on the image.

Mathematics
2 answers:
algol133 years ago
7 0

Answer:

10 + 10 = 21

Step-by-step explanation:

there there

quester [9]3 years ago
4 0

Answer:

4

Step-by-step explanation:

the 5 is thousand the 7 is ten thousand and the 4 is hundred thousand

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Which expression can be used to find the surface area of the following square pyramid? I DONT KNOW HOW TO PUT A PHOTO ON PC!!! b
chubhunter [2.5K]

Answer:

40 or 16+6+6+6+6

Step-by-step explanation:

To find the surface area of a 3d figure, we can imagine all of its faces laid down on a flat plane. In this case, we would have a square, and four congruent triangles. Now all we have to do is find the areas of each shape and add them up.

4 is the base of the pyramid, so it's also the square's side length. Since a square has four equal sides, our square's length and width are both 4.

4*4 = 16

For every triangle we have, the base is 4 and the height is 3. The area of a triangle can be found using the formula A=(bh)/2. We plug in the values:

A = (4*3)/2

A = (12)/2

A = 6

Since we have 4 triangles, the surface area is:

16+6+6+6+6 = 40

7 0
4 years ago
Find the slope of (-2,1),(2,-2)
Volgvan

Answer:

-3/4 is the slope.

Step-by-step explanation:

Use this formula to find slope.

y2-y1/x2-x1

PLUG IN

-2 -1/2- (-2) = -3/4

The slope is -3/4

3 0
4 years ago
Read 2 more answers
Find the areas of the sectors formed by ABC
Amanda [17]

Answer:

33.51 in^2

Step-by-step explanation:

Find the area of the circle:

A = πr^2

A = π * 8^2

A = 64π = 201.06 in^2

Find the area of the 60 deg slice:

A * 60/360 = 33.51 in^2

Leave a thanks and mark brainliest if this helped

Leave a thanks and mark brainliest if this helped

6 0
3 years ago
What is 1 tenth of 2.0
timofeeve [1]
O.2 0.2 0.2 the answer is 0.2
yeah
6 0
3 years ago
Read 2 more answers
Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane
liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

Answer:

Rate = 0.935042^\circ /cm

Step-by-step explanation:

Given

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

T(x,y) =x\sin2y

r = 1m

v = 2m/s

Express the given point P as a unit tangent vector:

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

u = \frac{\sqrt 3}{2}i - \frac{1}{2}j

Next, find the gradient of P and T using: \triangle T = \nabla T * u

Where

\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})}  = (sin \sqrt 3)i + (cos \sqrt 3)j

So: the gradient becomes:

\triangle T = \nabla T * u

\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] *  [\frac{\sqrt 3}{2}i - \frac{1}{2}j]

By vector multiplication, we have:

\triangle T = (sin \sqrt 3)*  \frac{\sqrt 3}{2} - (cos \sqrt 3)  \frac{1}{2}

\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)

\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5

\triangle T = 0.935042

Hence, the rate is:

Rate = \triangle T = 0.935042^\circ /cm

3 0
3 years ago
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