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dezoksy [38]
3 years ago
7

What is the greatest common factor of 8x and 40

Mathematics
1 answer:
Novay_Z [31]3 years ago
6 0
The GCF of 8x and 40 is 8
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Find the value of n in this equation.
lisabon 2012 [21]

Answer:

  • D. 9

=========================

<h2>Given </h2>

Triangle with:

  • Base of n² -3,
  • Midsegment of 39.

<h2>To find</h2>

  • The value of n.

<h2>Solution</h2>

As per definition of midsegment, it is connecting the midpoints of two sides and its length is half the length of the opposite side of the triangle.

So we have:

  • (n² - 3)/2 = 39

Solve it for n:

  • n² - 3 = 78
  • n² = 81
  • n = √81
  • n = 9

Correct choice is D.

3 0
1 year ago
Round each whole number to the given place value.
Kazeer [188]

We have to round each whole number to the given place value.

We have to round the number 437 to the tens place.

Since the digit at the tens place is 3.

Since the digit at its previous place that is ones place is '7' which is greater than 5.

Therefore, we can round the number 437 to 440.

Now, consider the another number 64,328, we have to round this number to the nearest ten thousand place.

Since the digit at ten thousands place is 6.

And the digit at its previous place that is, at thousands place is 4.

Since this number is less than 5.

Therefore, the given number will round to 60,000.

4 0
4 years ago
Sheniah went to McDonald's and bought a soda for one dollar. She also bought a Big Mac. For any price of the Big Mac, what would
Natalka [10]

Answer:

$1+x(price of big mac)

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?
Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

A(-6, 2)) = (x_1, y_1)

B(2, 6) = (x_2, y_2)

AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}

AB = \sqrt{(8)^2 + (4)^2}

AB = \sqrt{64 + 16} = \sqrt{80}

AB = 8.9 (nearest tenth)

Distance between B(2, 6) and C(7, 1):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

B(2, 6) = (x_1, y_1)

C(7, 1) = (x_2, y_2)

BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}

BC = \sqrt{(5)^2 + (-5)^2}

BC = \sqrt{25 + 25} = \sqrt{50}

BC = 7.1 (nearest tenth)

Distance between C(7, 1) and D(3, -5):

CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

C(7, 1) = (x_1, y_1)

D(3, -5) = (x_2, y_2)

CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}

CD = \sqrt{(-4)^2 + (-6)^2}

CD = \sqrt{16 + 36} = \sqrt{52}

CD = 7.2 (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

D(3, -5) = (x_1, y_1)

A(-6, 2) = (x_2, y_2)

DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}

DA = \sqrt{(-9)^2 + (7)^2}

DA = \sqrt{81 + 49} = \sqrt{130}

DA = 11.4 (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0
3 years ago
Select one of the factors of 5x2 + 7x + 2.
Dafna1 [17]
5x² + 5x + 2x + 2
5x(x + 1) + 2(x + 1)

(5x + 2)(x + 1)
6 0
3 years ago
Read 2 more answers
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