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Sidana [21]
3 years ago
10

Write the number 347.85 in expanded form. a 3 + 100 + 4 x 10 + 7 x 1 x 8 x (1/100) + 5 x (1/100) b 3 x 100 + 4 + 10 + 7 x 1 + 8

x (1/100) + 5 x (1/10) c 3 x 100 + 4 x 10 + 7 x 1 + 8 x (1/10) + 5 x (1/100)
Mathematics
2 answers:
SVEN [57.7K]3 years ago
8 0
The answer will be C
Andrews [41]3 years ago
6 0

Answer:

C. 3 × 100 + 4 × 10 + 7 × 1 + 8 × (1/10) + 5 × (1/100)

Step-by-step explanation:

3: hundreds

4: tens

7: ones

8: tenths

5: hundreths

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Y=2x^2+12x+14 in vertex form
balandron [24]

Answer:

y = 2(x + 3)² - 4

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Using the method of completing the square

y = 2x² + 12x + 14 ← factor out 2 from the first 2 terms

  = 2(x² + 6x) + 14

To complete the square

add/subtract ( half the coefficient of the x- term)² to x² + 6x

y = 2(x² + 2(3)x + 9 - 9 ) + 14

  = 2(x + 3)² - 18 + 14

  = 2(x + 3)² - 4 ← in vertex form

4 0
3 years ago
Which algebraic expression represents the perimeter of the rectangle?<br> 2x + 2<br> 4x + 3
Lapatulllka [165]
I’m pretty sure it’s the 3rd one
4 0
3 years ago
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E is the midpoint between D and F. If DE=3x+2, EF=5x, find the value of x. A. 1 B. 15 C. 3 D. 5
Nadya [2.5K]

Step-by-step explanation:

then u times the d and e to get the equivalent fraction to the number u get

7 0
3 years ago
QUESTION21 pointssave AnswerThe perimeter of a rectangle is 136 ft. The ratio of its length to its width is g 8, What are the di
jenyasd209 [6]
36ft by 32ft. I didn't do any math to solve this, but you can tell by comparing the ratios.

9:8 (Nine is just larger than eight, so you need to find an answer where the first number is just larger than the second. None of the others come anywhere close to that besides 36:32)
4 0
3 years ago
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Use the upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width)
Mashutka [201]
From the figure shown, the interval is divided into 5 equal parts making each subinterval to be 0.2.

Part A:

y= \sqrt{1-x^2}

The approximate the area of the region shown in the figure using the lower sums is given by:

 Area= [y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2]+[y(0.8)\times0.2] \\ +[y(1)\times0.2] \\  \\ =[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2]+[\sqrt{1-(0.6)^2}\times0.2] \\ +[\sqrt{1-(0.8)^2}\times0.2]+[\sqrt{1-(1)^2}\times0.2] \\  \\ =(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2)+(0\times0.2) \\  \\ =0.196+0.183+0.16+0.12=0.659



Part B:

The approximate the area of the region shown in the figure using the lower sums is given by:

 Area= [y(0)\times0.2]+[y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2] \\ +[y(0.8)\times0.2] \\ \\ =[\sqrt{1-(0)^2}\times0.2]+[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2] \\ +[\sqrt{1-(0.6)^2}\times0.2] +[\sqrt{1-(0.8)^2}\times0.2] \\ \\ =(1\times0.2)+(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2) \\ \\ =0.2+0.196+0.183+0.16+0.12=0.859



Part C:

The approximate area of the given region is given by

Area= \frac{0.659+0.859}{2} = \frac{1.518}{2} =0.759
7 0
4 years ago
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