Can someone please help me with all the question up to 49

Mathematics

1 answer:

Igoryamba2 years ago

4 0

Solve the top, then solve the bottom, and divide the top by the bottom.

You might be interested in

Find an equation for the linear function g(x) which is perpendicular to the line 5x-2y=6 and intersects the line 5x-2y=6 at x=12

exis [7]

Answer:

≡

Step-by-step explanation:

8 0

2 years ago

Rewrite in Polar Form....<br> x^(2)+y^(2)-6y-8=0

skad [1K]

$\bf x^2+y^2-6y-8=0\qquad 
\begin{cases}
x^2+y^2=r^2\\\\
y=rsin(\theta)
\end{cases}\implies r^2-6rsin(\theta)=8
\\\\\\
\textit{now, we do some grouping}\implies [r^2-6rsin(\theta)]=8
\\\\\\\
[r^2-6rsin(\theta)+\boxed{?}^2]=8\impliedby 
\begin{array}{llll}
\textit{so we need a value there to make}\\
\textit{a perfect square trinomial}
\end{array}$

$\bf 6rsin(\theta)\iff 2\cdot r\cdot \boxed{3\cdot sin(\theta)}\impliedby \textit{so there}
\\\\\\
\textit{now, bear in mind we're just borrowing from zero, 0}
\\\\
\textit{so if we add, \underline{whatever}, we also have to subtract \underline{whatever}}$

$\bf [r^2-6rsin(\theta)\underline{+[3sin(\theta)]^2}]\quad \underline{-[3sin(\theta)]^2}=8\\\\
\left. \qquad \right.\uparrow \\
\textit{so-called "completing the square"}
\\\\\\\
[r-3sin(\theta)]^2=8+[3sin(\theta)]^2\implies [r-3sin(\theta)]^2=8+9sin^2(\theta)
\\\\\\
r-3sin(\theta)=\sqrt{8+9sin^2(\theta)}\implies r=\sqrt{8+9sin^2(\theta)}+3sin(\theta)$

3 0

2 years ago

Can someone show me how to do this? I’m not sure how to do this

Allushta [10]

Answer: Slope means the gradient. So find the equation of the line first then graph the points on the graph.

Step-by-step explanation:

y=mx+c *We want m*

m=(y2-y1)/(x2-x1)

m=(0-1)/(6-3)

m=-1/3

y=-1/3x+2 *(0,2) co-ordinate means it's the y-intercept=c*

I think the slope is -1/3 and then just plt the points on the graph manually.

4 0

2 years ago

The problem is in the picture

creativ13 [48]

It should be the first answer choice between those two

6 0

3 years ago

Express y=3x+2 in general form

katrin [286]

General form of a linear equation is ax+by+c=0 where a,b, and c are three real numbers.

So, we can shift all the terms in one side of the equation. Hence, first step is to remove y from the left side. So, subtract y from each sides.

y-y=3x+2-y

0= 3x - y +2 (By simplifying)

So, the general form of the given equation is 3x-y+2=0.

4 0

3 years ago