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3 years ago

How do you determine whether line segment are congruent

Mathematics

2 answers:

svlad2 [7]3 years ago

6 0

If a line is congruent if it is doesn't have an opening or a curve

vovangra [49]3 years ago

5 0

They will usually have a dash in the line, or the same number next to both of them.

Hope this helped!

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Which number is greatest? 2.89 times 10 Superscript negative 8 1.997 times 10 Superscript 2 8.9 times 10 Superscript negative 6

love history [14]

<h3> The greatest number is $1.997 \times 10^2$ </h3>

Solution:

Given that the numbers are:

$2.89 \times 10^{-8}\\\\1.997 \times 10^2\\\\8.9 \times 10^{-6}\\\\5 \times 10^{-6}$

We have to find the number that is greatest

Convert the numbers to decimal

$2.89 \times 10^{-8} = 2.89 \times \frac{1}{10^8} =0.0000000289$

$1.997 \times 10^2 = 199.7$

$8.9\times 10^{-6}=8.9\times \frac{1}{10^6}=0.0000089$

$5\times 10^{-6}=5\times \frac{1}{10^6}=0.000005$

We can clearly see that, 199.7 is greatest number

Thus the greatest number is $1.997 \times 10^2$

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4 years ago

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Hey guys it's been a while can you just help me with these two problems?

lutik1710 [3]

Answer:

Your already right on both of them lol

Step-by-step explanation:

Pls brainliest thx! :D

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3 years ago

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Cedric's favorite popcorn cost $0.34 cents per ounce.How much would he speand on 17 ounces of popcorn?

Olin [163]

Answer:

$5.78

Step-by-step explanation:

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2 years ago

Let f(x) = |x| for all real numbers x. Write the formula for the function represented by the described

Ostrovityanka [42]

Answer:

$y=(\frac{1}{3}f(x-3))-1$

Step-by-step explanation:

Please see the picture below.

1. Given the function f(x) = |x|, applying a vertical stretch with scale factor $\frac{1}{3}$ , we have the transformed function:

$y=\frac{1}{3}f(x)$

2. Applying a translation of 3 units to the right, we have:

$y=\frac{1}{3}f(x-3)$

3. Finally applying a translation down of 1 unit, we have:

$y=(\frac{1}{3}f(x-3))-1$

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4 years ago

3/16 + 5/8 x 1/2 = A) 10/32 B) 5/16 C) 1/2 D) 1/6

madam [21]

0.5 in the pemdas form

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3 years ago

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