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Alina [70]
3 years ago
14

Use the model to write an equivalent fractions

Mathematics
2 answers:
Aleonysh [2.5K]3 years ago
8 0

Answer:4/6 = 2/3

Divide both by two

siniylev [52]3 years ago
7 0
These are both equivalent to 1

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7045 ÷ 15 is what? i can't do it
Kazeer [188]

Answer:

As a decimal, 469.666667. As a remainder, 269 r10. As a fraction,

469 with  666667 over  1000000

Step-by-step explanation:

Use a calculator and mental mat, I'm not great at explaining things.

Hope I could help though!

7 0
3 years ago
Read 2 more answers
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
Solve the following quadratic equation using the quadratic formula. Which of the following expressions gives the numerators of t
Illusion [34]

I hope the choices for the numerators of the solutions are given.

I am showing the complete work to find the solutions of this equation , it will help you to find an answer of your question based on this solution.

The standard form of a quadratic equation is :

ax² + bx + c = 0

And the quadratic formula is:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

So, first step is to compare the given equation with the above equation to get the value of a, b and c.

So, a = 10, b = -19 and c = 6.

Next step is to plug in these values in the above formula. Therefore,

x=\frac{(-19)-\pm\sqrt{(-19)^2-4*10*6}}{2*10}

=\frac{19\pm\sqrt{361-240}}{20}

=\frac{19\pm\sqrt{121}}{20}

=\frac{19\pm11}{20}

So, x=\frac{19-11}{20} ,\frac{19+11}{20}

x=\frac{8}{20} , \frac{30}{20}

So, x= \frac{2}{5} ,\frac{3}{2}

Hope this helps you!

5 0
3 years ago
Read 2 more answers
Circle O, with center (x, y) passes through the points A(0, 0), B(-3, 0), and C(1, 2). Find the coordinates of the center of the
KonstantinChe [14]

Answer:

The center of the circle is

(-\frac{3}{2},2)

Step-by-step explanation:

Let the equation of the circle be

x^2+y^2+2ax+2by+c=0, where (-a,-b) is the center of this circle.

The points lying the circle must satisfy the equation of this circle.

A(0,0)

We substitute this point to get;

0^2+0^2+2a(0)+2b(0)+c=0

\implies c=0

B(-3,0)

(-3)^2+0^2+2a(-3)+2b(0)+c=0

\implies 9+0-6a+0+c=0

\implies -6a+c=-9

But c=0

\implies -6a=-9

\implies a=\frac{3}{2}

C(1,2)

1^2+2^2+2a(1)+2b(2)+c=0

1+4+2a+4b+c=0

2a+4b+c=-5

Put the value of 'a' and 'c' to find 'b'

2(\frac{3}{2})+4b+0=-5

3+4b+0=-5

4b=-5-3

4b=-8

b=-2

Hence the center of the circle is

(-\frac{3}{2},2)

8 0
3 years ago
What is the slope between the points (-2,4) and (-3,5)
defon

Answer:

y = \frac{1}{-1}x+2

3 0
2 years ago
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