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wolverine [178]
3 years ago
6

I have more question just don't get it wronge

Mathematics
2 answers:
Phantasy [73]3 years ago
8 0
A


Mark brainliest please



Hope this helps you
lesya [120]3 years ago
4 0

Answer:

Relfection over the x axis

Step-by-step explanation:

It is the name shape but flipped onto the other side, therefore it is a reflection

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Why are halves a good choice for benchmark fractions for 1 1/3
Natalka [10]

Answer and explanation:

Benchmark fractions are fractions that are used as references in measuring other fractions. They are easily estimated and so can be used in measuring more "specific" fractions such as 1/5, 7/9, 3/7, 1/3 etc. If I wanted to measure 1 1/3cm for instance using a calibrated ruler, having centimeter measurements, I would first find 1cm on the ruler and then find half of one centimeter. Seeing that half is bigger than 1/3 but close, I could then estimate 1/3 to be somewhere less than 1/2 but a bit close to it

3 0
3 years ago
The measure of angle BFE is 120°. Find M angle EFD.
Varvara68 [4.7K]
120 + 120 + X + X = 360
240 + 2X = 360
2X = 120
X = 60
6 0
3 years ago
Please help and thank you. ​
Marina CMI [18]

Answer:

4.

  • 400 ft elevation at 1 km and 3 km from the start
  • 300 ft/km average rate of change over the first 4 km

5.

  • length = 3.5 cm
  • width = 3.5 cm
  • height = 7.0 cm

6.

  • 6 mo: $1018.14
  • 5 yrs: $1196.89
  • avg incr: $3.28 per month

Step-by-step explanation:

<h3>4.</h3>

A graphing calculator is handy for solving problems involving cubic polynomials. You're interested in where the elevation is 400 ft. Since the value of f(x) is in hundreds of feet, you want to find x such that f(x) = 4.

Values of x where that is the case are x=1 and x=3, representing distances of 1 km and 3 km from the start of the road.

The average rate of change of elevation is the difference in elevation divided by the difference in distance from the start:

  average rate of change = (f(4) -f(0) hundred ft)/((4 - 0) km)

  = (19 -7)/4 hundred feet/km

  = 12/4  hundred feet/km = 300 ft/km

___

<h3>5.</h3>

You want to find x when ...

  A(x) = 122.5 cm²

  10x² = 122.5 cm² . . . . substitute the given expression for A(x)

  x² = 12.25 cm² . . . . . . divide by 10

  x = √(12.25 cm²) = 3.5 cm . . . . take the square root

The diagram tells you ...

  length = width = x = 3.5 cm

  height = 2x = 7.0 cm

___

<h3>6.</h3>

Evaluate the given expression for the different values of m:

  $1000·1.003^6 ≈ $1018.14 . . . . 6-month value

  $1000·1.003^60 ≈ $1196.89 . . . . 5-year value

The increase is $1196.89 -1000.00 = $196.89. That increase took place over 60 months, so the average increase per month is ...

  $196.89/(60 mo) ≈ $3.28 per mo . . . . average per month over 5 years

8 0
4 years ago
Imagine you are filling up a water jug and the water is coming in at rate of 2 liters per 25 seconds. At this rate, how much wat
Zinaida [17]

The answer is 10.8  liters of water

Hope this helps!

4 0
3 years ago
If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
3 years ago
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