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Scorpion4ik [409]
3 years ago
12

One and one eighth equal g minus four and two fifths

Mathematics
1 answer:
Maurinko [17]3 years ago
5 0
1 1/8 = g - 4 2/5
9/8 = g - 22/5
9/8 + 22/5 = g
45/40 + 176/40 = g
221/40 = g
5 21/40 = g
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The wading pool shown is a trapezoidal prism with a total volume of 286 cubic feet. What is the missing dimension? First base: 8
harkovskaia [24]
To find the missing dimension you will use the formula for finding the volume of a prism and solve for the missing dimension.  In this case 2ft is how deep the pool is and you will solve for the width of the pool.

V = Bh, where B is the area of the base.  In this case, use the area of a trapezoid formula.

286 = 1/2h(8 +13)2
<u>286</u> = <u>21h</u>
21      21
h = 13.6

The width of the wading pool is approximately 13.6 feet.
5 0
3 years ago
Write 49,000 in scientific notation?
prisoha [69]

Answer:

4.9x10^4

Step-by-step explanation:

4.9*10000=49,000

8 0
3 years ago
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One of the vertices of an equilateral triangle is on the vertex of a square and two other vertices are on the not adjacent sides
Elina [12.6K]
<h2>Answer:</h2>

<em> The side of the triangle is either 38.63ft or 10.35ft</em>

<h2>Step-by-step explanation:</h2>

This problem can be translated as an image as shown in the Figure below. We know that:

  • The side of the square is 10 ft.
  • One of the vertices of an equilateral triangle is on the vertex of a square.
  • Two other vertices are on the not adjacent sides of the same square.

Let's call:

Since the given triangle is equilateral, each side measures the same length. So:

x: The side of the equilateral triangle (Triangle 1)

y: A side of another triangle called Triangle 2.

That length is the hypotenuse of other triangle called Triangle 2. Therefore, by Pythagorean theorem:

\mathbf{(1)} \ x^2=100+y^2

We have another triangle, called Triangle 3, and given that the side of the square is 10ft, then it is true that:

y+(10-y)=10

Therefore, for Triangle 3, we have that by Pythagorean theorem:

(10-y)^2+(10-y)^2=x^2 \\ \\ 2(10-y)^2=x^2 \\ \\ \\ \mathbf{(2)} \ x^2=2(10-y)^2

Matching equations (1) and (2):

2(10-y)^2=100+y^2 \\ \\ 2(100-20y+y^2)=100+y^2 \\ \\ 200-40y+2y^2=100+y^2 \\ \\ (2y^2-y^2)-40y+(200-100)=0 \\ \\ y^2-40y+100=0

Using quadratic formula:

y_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ y_{1,2}=\frac{-(-40) \pm \sqrt{(-40)^2-4(1)(100)}}{2(1)} \\ \\ \\ y_{1}=37.32 \\ \\ y_{2}=2.68

Finding x from (1):

x^2=100+y^2 \\ \\ x_{1}=\sqrt{100+37.32^2} \\ \\ x_{1}=38.63ft \\ \\ \\ x_{2}=\sqrt{100+2.68^2} \\ \\ x_{2}=10.35ft

<em>Finally, the side of the triangle is either 38.63ft or 10.35ft</em>

5 0
3 years ago
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What's the rate of -3(x-2)-8
Charra [1.4K]

Answer:

-3x-2

Step-by-step explanation:

5 0
2 years ago
Help me with this please!!!!!!!!!
IrinaVladis [17]

Answer:

d = 17.58

Step-by-step explanation:

d = √(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2

P1(0, 0, 0) P2(8, 7, 14)

Substitute the values in the above equation and you will get

d = √(8 - 0)^2 + (7 - 0)^2 + (14 - 0)^2

= √(64 + 49 + 196)

= √309

= 17.58

8 0
3 years ago
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