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Leya [2.2K]
3 years ago
13

Turn 9/4 into a mixed number

Mathematics
2 answers:
natta225 [31]3 years ago
3 0

2 1/4 you divide 4 into nine and minus the remainder

Nonamiya [84]3 years ago
3 0

Answer:

2 1/4

Step-by-step explanation:

Two wholes can be taken from the 9/4, leaving you with 1/4, and you would write that as 2 1/4

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Square ABCD and isosceles triangle BUC are drawn to create trapezoid AUCD. Square A B C D and triangle B U C are attached at sid
Art [367]

Answer:

d) 135º

Step-by-step explanation:

Note that the angle DCU is the sum of the angles DCB and BCU. The angle DCB is 90º because A B C D is a square, then all its angles are equal to 90º.

After attaching B U C to A B C D, we obtain a trapezoid A U C D. Since A U C D has at least one pair of parallel sides, then AU should be parallel to CD, thus the angle CBU must be 90º.

B U C is isoceles, so we conclude that other two angles must have the same size, and due to the sum of the angles of a triangle being 180º, then both BUC and BCU are equal to 45º

As a result, the angle DCU is equal to 90º+45º = 135º. Option d is the correct one.

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4 years ago
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Factor 15y2 + 10y − 40.
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3 years ago
Which transformation carries the pentagon onto its
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6 0
3 years ago
Nelda drove 315 kilometers and used 35 liters of gasoline. Her car averaged how many kilometers per liter of gas?​
Inessa [10]

Going 315 km uses up 35 liters of gasoline

On average, 1 liter of gasoline can travel the number of kilometers:

315 : 35 = 9 liters of gasoline

6 0
3 years ago
Let f(x)=x^3-x-1
alexira [117]
f(x) = x^3 - x - 1

To find the gradient of the tangent, we must first differentiate the function.

f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1

The gradient at x = 0 is given by evaluating f'(0).

f'(0) = 3(0)^2 - 1 = -1

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

y = -x + c

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

f(0) = (0)^3 - (0) - 1 = -1

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}
6 0
3 years ago
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