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denpristay [2]
3 years ago
13

9.014 which digit is in the hundredths place

Mathematics
2 answers:
Crank3 years ago
5 0

Answer:

1/100 AKA 1.

Step-by-step explanation:

O.T (H) T

9. 0 1  4

jekas [21]3 years ago
4 0

Answer:

Step-by-step explanation:

1 is in the hundredths

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The blueprint for the Moreno Moreno​'s living room has a scale of 2 inches equals 7 feet 2 inches=7 feet. The family waThe bluep
lubasha [3.4K]

<u>Corrected Question</u>

The blueprint for the Moreno Moreno​'s living room has a scale of 2 inches equals 7 feet.The family wants to use a scale of 1 inch equals = 6 feet. What is the width of the living room on the new​ blueprint? A rectangle is labeled width = 12 inches.

Answer:

7 Inch

Step-by-step explanation:

The width of the room on the old blueprint = 12 Inches

<u>Scale on the old blueprint: </u>

2 inches = 7 feet.

Therefore: 1 Inch =7/2 feet

12 Inches =12 X 7/2 feet =42 feet

The actual width of the room is 42 feet.

<u>Scale on the new blueprint</u>

1 inch equals = 6 feet; which we can reorder as:

6 feet=1 inch

1 feet =1/6 Inch

Therefore:

42 feet = 42 x 1/6 Inch =7 Inch

Thus, the width of the living room on the new​ blueprint is 7 Inch.

7 0
3 years ago
Using the numbers -4, 10, 8, 2, -3, -5 create two expressions that equal six
docker41 [41]

Answer:

8-2 and 10-3

Step-by-step explanation:

3 0
2 years ago
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The graph of a quadratic function intercepts the x-axis in the two places and the y-axis in one place.
wolverine [178]
Correct Answer:
Option 3: <span>The quadratic function has two distinct real zeros.

The function is quadratic, therefore it can have only 2 zeros. The knowledge of x-intercepts is needed to determine the zeros, y-intercepts has nothing to do with the zeros of a function. The given function has 2 unique x-intercepts, so according to the fundamental theorem of algebra, this function has 2 distinct real roots as number of distinct real roots are equal to the number of x-intercepts. Therefore, option 3 is the correct answer. </span>
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Which set of numbers could represent the lengths of the sides of a right triangle?
DiKsa [7]

Answer:

The first set: 8, 15, and 17.

Step-by-step explanation:

<h3>Pair: 8, 15, 17</h3>

By the pythagorean theorem, a triangle is a right triangle if and only if

\text{longest side}^2 = \text{first shorter side}^2 + \text{second shorter side}^2.

In this case,

\text{longest side}^2 = 17^2 = 289.

\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 8^2 + 15^2\\ &=64 + 225 = 289 \end{aligned}.

In other words, indeed \text{hypotenuse}^2 = \text{first leg}^2 + \text{second leg}^2. Hence, 8, 15, 17 does form a right triangle.

Similarly, check the other pairs. Keep in mind that the square of the longest side should be equal to the sum of the square of the two

<h3>Pair: 10, 15, 20</h3>

Factor out the common factor 2 to simplify the calculations.

\text{longest side}^2 = 20^2 = 400

\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 10^2 + 15^2\\ &=100 + 225 = 325 \end{aligned}.

\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2.

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

<h3>Pair: 12, 18, 22</h3>

\text{longest side}^2 = (2\times 11)^2 = 2^2 \times 121.

\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= (2 \times 6)^2 + (2 \times 9)^2\\ &=2^2 \times(36 + 81) = 2^2 \times 117 \end{aligned}.

\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2.

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

<h3>Pair: 7, 9, 11</h3>

\text{longest side}^2 = 11^2 = 121.

<h3>\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 7^2 + 9^2\\ &=49+ 81 = 130 \end{aligned}.</h3>

\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2.

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

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3 years ago
One student ate 3/20 of all candies and another 1.2 lb. The second student ate 3/5 of the candies and the remaining 0.3 lb. What
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The students ate 6 lbs of candies.
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