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3 years ago

Can you help me with this question please

Mathematics

1 answer:

vaieri [72.5K]3 years ago

5 0

To get the answer, divide 55 by 610. the answer would be 11.09 i’m pretty sure

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Find the sum of the digits 10^15−1.

Alchen [17]

Answer:

149

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Why is this correct?

Vanyuwa [196]

Answer: A= independent variable W = dependent variable

Step-by-step explanation:

w= a/25

7 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,

$\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}$

The answer is none of the choices

7 0

1 year ago

Greg is trying to solve a puzzle where he has to figure out two numbers, x and y. Three less than two-third of x is greater than

Fittoniya [83]

Inequation 1:

$\frac{2}{3}x-3 \geq y$

to plot the pairs (x, y) for which the inequation holds, draw the line $y=\frac{2}{3}x-3$

then pick a point in either side of the line. If that point is a solution of the inequation, than color that region of the line, if that point is not a solution, then color the other part of the line.

we do the same for the second inequation. Then the solution, is the region of the x-y axes colored in both cases.

inequation 2:

$y+ \frac{2}{3}x\ \textless \ 4$

$y\ \textless \ - \frac{2}{3} x+ 4
$

draw the lines

i) $y=\frac{2}{3}x-3$ use points (0, -3), (3, -1)

ii) $y=- \frac{2}{3} x+ 4$ use points ( 0, 4), (3, 2)

let's use the point P(3, 3) to see what region of the lines need to be coloured:

$\frac{2}{3}x-3 \geq y$ ;
$\frac{2}{3}(3)-3 \geq 3$
$2-3 \geq 3$ , not true so we color the region not containing this point

$y+ \frac{2}{3}x\ \textless \ 4$
$(3)+ \frac{2}{3}(3)\ \textless \ 4$
$3+ 5\ \textless \ 4$ not true, so we color the region not containing the point (3, 3)

The graph representing the system of inequalities is the region colored both red and blue, with the blue line not dashed, and the red line dashed.

4 0

3 years ago

HELP!

Advocard [28]

9)its a fraction

x=−451/19

y=−302/19

16)

Line 1: y = 5x +13

Line 2: y = 5x + 12

7 0

3 years ago