Answer:
8
Step-by-step explanation:
Considering the given value of sum of interior angles with exclusion of one to be 1070 then we can write it as
1070=180(n-2) where n is the number of sides required here.
Making n the subject of the formula then following step by step
![n-2=\frac {1070}{180}](https://tex.z-dn.net/?f=n-2%3D%5Cfrac%20%7B1070%7D%7B180%7D)
![n=\frac {1070}{180}+2\\n=7.9444444444444\approx 8](https://tex.z-dn.net/?f=n%3D%5Cfrac%20%7B1070%7D%7B180%7D%2B2%5C%5Cn%3D7.9444444444444%5Capprox%208)
Therefore, the shape is octagon and it has 8 sides.
Since we know y=5x - 3 and we know that x=0 then its a matter of plugging in what we know and simplifying. y=5 x 0 - 3 we need to remember the order of operations for this problem. PEMDAS parenthesis exponent multiply divide add subtrac. Since 5 time 0 is multiplication and multiplication comes before subtracting in the oder of operations we will solve that first. 5 times 0 equals 0 since anything times 0 is 0. All that's left is to subtract 3 which gives us y=-3 <span />
Answer:
![-5-\sqrt{-44}=-5-2\sqrt{11}i](https://tex.z-dn.net/?f=-5-%5Csqrt%7B-44%7D%3D-5-2%5Csqrt%7B11%7Di)
Step-by-step explanation:
We want to simplify:
![-5-\sqrt{-44}](https://tex.z-dn.net/?f=-5-%5Csqrt%7B-44%7D)
We rewrite the expression under the radical sign to obtain:
![-5-\sqrt{4\times11\times -1}](https://tex.z-dn.net/?f=-5-%5Csqrt%7B4%5Ctimes11%5Ctimes%20-1%7D)
We split the expression under the radical sign to get:
![-5-\sqrt{4} \times \sqrt{11} \times \sqrt{-1}](https://tex.z-dn.net/?f=-5-%5Csqrt%7B4%7D%20%5Ctimes%20%5Csqrt%7B11%7D%20%5Ctimes%20%5Csqrt%7B-1%7D)
Recall that:
![\sqrt{-1}=i](https://tex.z-dn.net/?f=%5Csqrt%7B-1%7D%3Di)
This implies that:
![-5-\sqrt{4} \times \sqrt{11} \times \sqrt{-1}=-5-2\sqrt{11}i](https://tex.z-dn.net/?f=-5-%5Csqrt%7B4%7D%20%5Ctimes%20%5Csqrt%7B11%7D%20%5Ctimes%20%5Csqrt%7B-1%7D%3D-5-2%5Csqrt%7B11%7Di)
Therefore ![-5-\sqrt{-44}=-5-2\sqrt{11}i](https://tex.z-dn.net/?f=-5-%5Csqrt%7B-44%7D%3D-5-2%5Csqrt%7B11%7Di)
Answer:
-5 < x < 5
Step-by-step explanation:
|x| < 5
This can be split into two cases: x < 5 and -x < 5 because |x| could be either x or -x depending on whether x is positive, negative, or 0. Solving the second one we get x > -5 so the answer is -5 < x < 5.
Applying the rule of modulus for given values, the expression will result to 20.
<h3>What is Modulus or Absolute Values</h3>
The modulus of a value for example |a|=a if a is greater than or equal to zero
and also
|a|=-a if a is less than zero
In the expression given; the modulus of -6 written as
|-6|=-(-6) this is because the value is less than zero
The modulus of 2 written as; |2|=2 this is because the value is greater than zero. and;
|The modulus of -14 written as; |-14|= -(-14) since the value is less than zero.
Hence, we can rewrite the expression as;
we deal with multiplication first following the rules of BODMAS
2×-(-6)-3×2+[-(-14)]
2×6-6+14
and thus adding values
12+14-6
and finally we subtract;
26-6
which will results to the solution of 20.