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xxTIMURxx [149]
3 years ago
5

please solve the question is this 176 x 17 and 180 x 20 = 3600 overestimated or underestimated a. overestimated or b. underestim

ated make sure u show me why its the right answer and where your answer from
Mathematics
1 answer:
Citrus2011 [14]3 years ago
4 0
176 rounds up to 180
17 rounds up to 20
The fact we're rounding up both times means we'll have an overestimate
estimate = 180*20 = 3600
actual result = 176*17 = 2992
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The perimeter of a rectangle is 50 centimeters. If the width of the rectangle is 9 centimeters
nordsb [41]

Answer:

Length = 17 cm

Width = 8 cm

Step-by-step explanation:

Let the length be x cm

Therefore, Width = (x - 9) cm

Perimeter of rectangle = 2(l +w)

50 = 2[x + (x - 9)]

50/2 = 2x - 9

25 = 2x - 9

25+ 9 = 2x

2x = 34

x = 34/2

x = 17 cm

x - 9 = 17 - 9 = 8 cm

Therefore,

Length = 17 cm

Width = 8 cm

Thus, the dimensions of the rectangle are 17 cm and 8 cm.

6 0
3 years ago
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8 0
3 years ago
Expand using the properties and rules for logarithms
malfutka [58]

Consider expression \log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).

1. Use property

\log_a\dfrac{b}{c}=\log_ab-\log_ac.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.

2. Use property

\log_abc=\log_ab+\log_ac.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2.

3. Use property

\log_ab^k=k\log_ab.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2.

4. Use property

\log_{a^k}b=\dfrac{1}{k}\log_ab.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{2^{-1}}2=\\ \\=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+\log_22=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+1.

Answer: correct option is B.

7 0
4 years ago
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