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stiv31 [10]
3 years ago
11

12.5 feet long weighs 158 pounds. How much does a piece of the same steel pipe 18.75

Mathematics
1 answer:
AveGali [126]3 years ago
4 0

Answer:

237 pounds

Step-by-step explanation:

237 pounds

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The graph of

... y = 3/4 - 2

is a horizontal line at y = -1.25.

The graph of

... y = (3/4)x - 2

is an upward-sloping line with a rise of 3 for every 4 units to the right. It intercepts the y-axis at y = -2.

3 0
3 years ago
If each side of a square grows 6 mm, the area of ​​the square is multiplied by 16. How long are the sides of the original square
Phoenix [80]

Answer:

2 mm

Step-by-step explanation:

Let x be the sides of the original square,

then the area of the original square is x^2

The sides of the new square is x + 6

The area is (x + 6)^2

Then (x + 6)^2 = 16*x^2

Take square root on both sides and get

x + 6 = 4x,

3x = 6

so x = 2 mm

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3 years ago
Please a genius help me please
Vedmedyk [2.9K]

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B, D, and F

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2 years ago
Find the vertex and the axis of symmetry of the graph of y=-6(x+4)^2 – 3.
Serjik [45]
Vertex is (4,-3) and the axis of symmetry is X=4
4 0
3 years ago
PLEASE PLEASE ANSWER!
muminat

The area of the shaded region is $8(\pi \ -\sqrt{3})\ \text{cm}^2.

Solution:

Given radius = 4 cm

Diameter = 2 × 4 = 8 cm

Let us first find the area of the semi-circle.

Area of the semi-circle = \frac{1}{2}\times \pi r^2

                                      $=\frac{1}{2}\times \pi\times 4^2

                                      $=\frac{1}{2}\times \pi\times 16

Area of the semi-circle = $8\pi cm²

Angle in a semi-circle is always 90º.

∠C = 90°

So, ABC is a right angled triangle.

Using Pythagoras theorem, we can find base of the triangle.

AC^2+BC^2=AB^2

AC^2+4^2=8^2

AC^2=64-16

AC^2=48

AC=4\sqrt{3} cm

Base of the triangle ABC = 4\sqrt{3} cm

Height of the triangle = 4 cm

Area of the triangle ABC = \frac{1}{2}\times b \times h

                                          $=\frac{1}{2}\times 4\sqrt{3}  \times 4

Area of the triangle ABC =  8\sqrt{3} cm²

Area of the shaded region

                   = Area of the semi-circle – Area of the triangle ABC

                   = $8\pi \ \text{cm}^2-8\sqrt{3}\ \text{cm}^2

                   = $8(\pi \ -\sqrt{3})\ \text{cm}^2

Hence the area of the shaded region is $8(\pi \ -\sqrt{3})\ \text{cm}^2.

3 0
3 years ago
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