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Vanyuwa [196]
4 years ago
13

Is -0 greater then 0

Mathematics
2 answers:
pav-90 [236]4 years ago
5 0

Answer:

-0 is less than 0.

Explanation:

think about the number line

numbers on the left side are less means -negative

numbers on the right side are greater means- positive

DENIUS [597]4 years ago
3 0

Answer:

0 is greater than -0

Hope this helps

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Solve the equation.<br><br> 3<br> 4<br> (2a − 6) + 1<br> 2<br> = 2<br> 5<br> (3a + 20)<br><br> a =
zvonat [6]

Answer:

a = 40 if simplified :)

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The area of the base of the regular quadrilateral pyramid is 36 cm2 and the area of a lateral face is 48 cm2. Find the Lateral a
Mkey [24]
The lateral area is 192 cm².

The lateral area is the area of all of the lateral faces.  Since the base of the pyramid is a quadrilateral, there are 4 lateral faces.  The area of one lateral face is 48 cm²; this means the lateral area is 48(4) = 192 cm².
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3 years ago
Use the Exterior Angle Theorem to find the measure of each angle.
navik [9.2K]

Answer:

∠ C = 36°

∠ D= 81°

∠ DEC= 63°

Step-by-step explanation:

<em>The exterior angle of a triangle is equal to the sum of the interior opposite angles. </em>

<em />

∴   (11y + 4) + (5y + 1) = 117

         11y + 4 + 5y + 1 = 117

         16y + 5 = 117

         16y = 117-5

         16y = 112

         y = 112÷16

         y = 7

∠ C = 5×(7) + 1 = 36°

∠ D = 11×(7) + 4 = 81°

∠ DEC = 180- 117 = 63°

<u><em>PLEASE MARK THIS ANSWER AS THE BRAINLIEST</em></u>

<u><em></em></u>

<em />

7 0
3 years ago
A line contains the points (34, 12) and (32, 48) .
KiRa [710]
<span>slope intercept form is : y = mx + b your looking for the slope it is just (y-y/x-x)
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5 0
4 years ago
From the sum of a2−2ab+b2 and 2a2+2ab+b2 subtract the sum of a2−b2 and a2+ab+3b2
Sedaia [141]

Answer:

a^2-ab

Step-by-step explanation:

We need to find the sum of a^2-2ab+b^2 and 2a^2+2ab+b^2 first.

Adding a^2-2ab+b^2 + 2a^2+2ab+b^2

Combining like terms, we get

a^2+2a^2=3a^2

-2ab+2ab = 0

b^2+b^2=2b^2

Therefore,

a^2-2ab+b^2 +2a^2+2ab+b^2=3a^2+2b^2.

Now, we need to find the sum of a^2-b^2 \ \ and \ \ a^2+ab+3b^2..

Adding a^2−b^2+a^2+ab+3b^2.

Combining like terms, we get

a^2+a^2=2a^2

-b^2+3b^2=2b^2.

Therefore,

a^2−b^2+a^2+ab+3b^2=2a^2+ab+2b^2.

Now, subtracting

3a^2+2b^2 -(2a^2+ab+2b^2).

Distributing minus sign over second parenthesis, we get

3a^2+2b^2-2a^2-ab-2b^2.

Combining like terms,

3a^2-2a^2=a^2

2b^2-2b^2=0

Therefore,

3a^2+2b^2-2a^2-ab-2b^2=a^2-ab.

Therefore, the difference of the sum of a^2-2ab+b^2 + 2a^2+2ab+b^2 and a^2-b^2+a^2+ab+3b^2. is a^2-ab.

5 0
3 years ago
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