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lara [203]
3 years ago
12

A rectangle has dimensions 5 and one fourth

Mathematics
1 answer:
trapecia [35]3 years ago
3 0
Look at the picture, the triangles are half a rectangle. Just use the formula for quadrilateral and divide by two
5 \frac{1}{4}  \times 11 = 57.75
57.75÷2=28.875in²
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HELP PLS
olchik [2.2K]

3996  \times  \frac{1}{3}  = 1332
the volume is 1332 in^3


good luck
4 0
3 years ago
What times itself equals -10
lutik1710 [3]
0x 10 = -10
I think
7 0
3 years ago
7x(x+1.8)=0<br> 7x(x+1.8)=0<br> 7x(x+1.8)=0 <br> Answer those 3
ahrayia [7]

Answer:

The answer to all of those questions would be x=-1.8

Step-by-step explanation:

7x(x + 1.8) = 0 - Distribute through the Parenthesis -

7x^2 + 12.6x = 0

x(7x + 12.6) = 0


x = 0


7x + 12.6 = 0

7x = -12.6

x = -12.6/7

x = -1.8


so x = 0 or x = -1.8

3 0
3 years ago
The first thing you should do in solving a word problem is estimate what you think the answer would be.
Lelu [443]

Answer:

The answer is True.

Step-by-step explanation:

  • <u>Estimate the Answer Before Solving Having a general idea of a ballpark answer .</u>
  • <u>Or the problem lets students know if their actual answer is reasonable or not.</u>
6 0
2 years ago
The region bounded by y=(3x)^(1/2), y=3x-6, y=0
Ganezh [65]

Answer:

4.5 sq. units.

Step-by-step explanation:

The given curve is y = (3x)^{\frac{1}{2} }

⇒ y^{2} = 3x ...... (1)

This curve passes through (0,0) point.

Now, the straight line is y = 3x - 6 ....... (2)

Now, solving (1) and (2) we get,

y^{2} - y - 6 = 0

⇒ (y - 3)(y + 2) = 0

⇒ y = 3 or y = -2

We will consider y = 3.

Now, y = 3x - 6 has zero at x = 2.

Therefor, the required are = \int\limits^3_0 {(3x)^{\frac{1}{2} } } \, dx - \int\limits^3_2 {(3x - 6)} \, dx

= \sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}

= [\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} }  }{3}] - [13.5 - 18 - 6 + 12]

= 6 - 1.5

= 4.5 sq. units. (Answer)

7 0
3 years ago
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