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OLga [1]
3 years ago
10

Answer both questions please ! Question 7 D is cut off D. 2110.1

Mathematics
1 answer:
ANTONII [103]3 years ago
5 0
The answers are B and D
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Solve 240 = 6z + c for 'c'.<br><br> show steps please
Lunna [17]

There are two ways to find or determine for the value of c. In the first method, we can use addition and subtraction to isolate the variable c from the other variables. In the second method, we can use the transposition of variables to isolate the variable c from the other variables.

So solving for the value of c:

 

<span>Using 1st method: Addition and Subtraction</span>

We are given:

240 = 6 z + c

Simply subtract 6 z on both sides:

240 – 6 z = 6 z + c – 6 z

Cancelling 6 z – 6 z on the right side:

240 – 6 z = c

or

c = 240 – 6 z

 

<span>Using the 2nd method: Transposition</span>

240 = 6 z + c

What we are going to do here is to simply transpose the variable 6 z from the right side to the left side of the equation so that we are left with c alone on the right side. Always remember that when we transpose, the symbol becomes opposite. That is:

240 + (- 6 z) = c

240 – 6 z = c

or

<span>c = 240 – 6 z</span>

6 0
3 years ago
Given: tangent to Circle O.<br><br> If m C = 57°, then m BDR =
Romashka-Z-Leto [24]
∠BCD = 57°
∴ ∠BDR = ∠BCD = 57° (angle that meets the chord and the tangent is equi-angular to the angle at the alternate segment)
8 0
3 years ago
If A(-5,7), B(-4,-5), C(-1,-6) and D(4,5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.
MrMuchimi
After plotting the quadrilateral in a Cartesian plane, you can see that it is not a particular quadrilateral. Hence, you need to divide it into two triangles. Let's take ABC and ADC.

The area of a triangle with vertices known is  given by the matrix
M = \left[\begin{array}{ccc} x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right]

Area = 1/2· | det(M) |
        = 1/2· | x₁·y₂ - x₂·y₁ + x₂·y₃ - x₃·y₂ + x₃·y₁ - x₁·y₃ |
        = 1/2· | x₁·(y₂ - y₃) + x₂·(y₃ - y₁) + x₃·(y₁ - y₂) |

Therefore, the area of ABC will be:
A(ABC) = 1/2· | (-5)·(-5 - (-6)) + (-4)·(-6 - 7) + (-1)·(7 - (-5)) |
             = 1/2· | -5·(1) - 4·(-13) - 1·(12) |
             = 1/2 | 35 |
             = 35/2

Similarly, the area of ADC will be:
A(ABC) = 1/2· | (-5)·(5 - (-6)) + (4)·(-6 - 7) + (-1)·(7 - 5) |
             = 1/2· | -5·(11) + 4·(-13) - 1·(2) |
             = 1/2 | -109 |
<span>             = 109/2</span>

The total area of the quadrilateral will be the sum of the areas of the two triangles:

A(ABCD) = A(ABC) + A(ADC) 
               = 35/2 + 109/2
               = 72
8 0
3 years ago
Given the function f(x) = 2|x + 6| -4, for what values of x is f(x) = 6?
photoshop1234 [79]

Answer:

f(x) = 2|x + 6|-4

6=2|x + 6|-4

10= 2|x + 6|

divide by 2

5=|x + 6|

x = 5-6

x = -1

so x should be -1

4 0
3 years ago
Can you add and subtract integers and surds together?
Ne4ueva [31]

Answer:

Both of these examples are wrong. You cannot add/subtract integers and square roots together, however, you could add square roots together if they have the same number under the square root. For example, 2 - 2√6 will stay as 2 - 2√6 because they aren't like terms. 25 + 5√5 + 5√5 + 5 = 30 + 10√5 because 25 + 5 = 30 and 5√5 + 5√5 = 10√5. We can add 5√5 and 5√5 together because they have the same number under the square root. If we were to compute √2 + √3, we would just leave it as is because they don't have the same number under the square root.

4 0
3 years ago
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