Please help !

Mathematics

1 answer:

nadya68 [22]2 years ago

6 0

Answer:

r(-2)= -5

r(0)= -7

r(5)= -12

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Which function below has the lowest y intercept? f(x) graph going through (0, 5) and (2, 1) g(x) Kyle started the summer having

mr Goodwill [35]

First find the slope of f(x).

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

m=(1-5)/(2-0)

m=-4/2

m=-2

y=-2x+b, using (2,1) we can solve for the y-intercept, "b"

1=-2(2)+b

1=-4+b

5=b

y=-2x+5

So f(x) has a y-intercept of 5

g(x)=6m+3

So g(x) has a y-intercept of 3

h(x)=3x+4

So h(x) has a y-intercept of 4

Then g(x) has the lowest y-intercept of just 3.

6 0

3 years ago

Read 2 more answers

the rectangular grassy plot 126 m by 48m has a gravel path 1.5m wide a road 2.5m wide is constructed all around it on inside of

dusya [7]

Answer:

= Sh. 2281.50

Step-by-step explanation:

Area of the path;

(126 × 2.5 ) + (2.5 × (48 - (2.5×2)

= 315 + (2.5 × 43)

= 315 + 107.5

= 422.5 m²

But; the cost of graveling is sh 5.40 per square meter

Therefore; The cost of graveling the whole path;

= 422.5 × 5.40

<u>= Sh. 2281.50</u>

6 0

3 years ago

Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

Also,