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attashe74 [19]
3 years ago
8

I need help please help me

Mathematics
2 answers:
Elan Coil [88]3 years ago
6 0
Here’s the correct answer and how I got it :)

Mrac [35]3 years ago
4 0
Heyy I wanted to help but someone else answered
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What is the measure of angles 1, 2, and 3?​
stellarik [79]

Answer:

1=  70

2= 65

3=  95

For 1, just solve 180-(65+45) and then for 2, subtract #1 and 45 from 180. Then just solve for #3 by using 180-(#2+20)

4 0
3 years ago
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Write a linear function f with the values f(-4) = 0 and f(0) =0 please help
Sergeeva-Olga [200]

f(\stackrel{x_1}{-4})=\stackrel{y_1}{0}\qquad f(\stackrel{x_2}{0})=\stackrel{y_2}{0}~\hfill (\stackrel{x_1}{-4}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-4)}}}\implies \cfrac{0}{0+4}\implies \cfrac{0}{4}\implies 0

\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{0}(x-\stackrel{x_1}{(-4)})\implies \boxed{y = 0}

6 0
3 years ago
Solve for X
vovangra [49]

Answer:

308 degree

Step-by-step explanation:

because 360 subtract 52

6 0
3 years ago
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What is the solution for <br> x - 4 &lt; 3
Nikitich [7]
X < 7! just add 4 to both sides.
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3 years ago
If f(1) = 0, what are all the roots of the function f(x)=x^3+3x^2-x-3? Use the Remainder Theorem.
Sophie [7]
There's no if about it, 

f(x)=x^3+3x^2-x-3&#10;

has a zero f(1)=0 so x-1 is a factor.   That's the special case of the Remainder Theorem; since f(1)=0 we'll get a remainder of zero when we divide f(x) by x-1.

At this point we can just divide or we can try more little numbers in the function.  It doesn't take too long to discover f(-1)=0 too, so  x+1 is a factor too by the remainder theorem.  I can find the third zero as well; but let's say that's out of range for most folks.

So far we have 

x^3+3x^2-x-3 = (x-1)(x+1)(x-r)

where r is the zero we haven't guessed yet.  Again we could divide f(x) by (x-1)(x+1)=x^2-1 but just looking at the constant term we must have

-3 = -1 (1)(-r) = r

so

x^3+3x^2-x-3 = (x-1)(x+1)(x+3)

We check f(-3)=(-3)^3+3(-3)^2 -(-3)-3 = 0 \quad\checkmark

We usually talk about the zeros of a function and the roots of an equation; here we have a function f(x) whose zeros are

x=1, x=-1, x=-3

8 0
3 years ago
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