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Yuki888 [10]
3 years ago
7

Need some help with this math question thanks :)

Mathematics
1 answer:
Anton [14]3 years ago
3 0

Answer:

Option B \frac{24}{25}

Step-by-step explanation:

we know that

sin(2\theta) = 2 sin(\theta) cos(\theta)

In this problem we have

cos(\theta)=\frac{4}{5}

Find the value of sin(\theta)

Remember that

sin^{2} (\theta)+cos^{2} (\theta)=1

substitute the value

sin^{2} (\theta)+(\frac{4}{5})^{2}=1

sin^{2} (\theta)=1-\frac{16}{25}

sin^{2} (\theta)=\frac{9}{25}

sin(\theta)=\frac{3}{5}

Find the value of sin(2\theta)

sin(2\theta) = 2 (\frac{3}{5})(\frac{4}{5})

sin(2\theta) = \frac{24}{25}

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Find the gradient of the line segment between the points (-8,6) and (3,8).
sergey [27]

Answer:

Step-by-step explanation:

The gradient or slope of a line is defined as

m=(y2-y1)/(x2-x1), we have points (-8,6) and (3,8)

m=(8-6)/(3-(-8))

m=2/11

4 0
3 years ago
Plz, help me with this question!
Sever21 [200]

Answer:

<h3>a. 60 </h3>

Step-by-step explanation:

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4 0
3 years ago
Can someone please help me figure out what’s 18 3/4 as a decimal
dmitriy555 [2]

Answer:

18.75

Step-by-step explanation:

If you covert this into a mixed fraction it would be 75/4 (4*18 is 72 then add the 3 in the numerator) then divide, 75/4 = 18.75 (sorry if explanation is bad)

7 0
3 years ago
Read 2 more answers
Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!
IgorC [24]
Hey there, hope I can help!

-4x^2+9y^2+32x+36y-64=0

\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \  9y^2+32x+36y-4x^2=64

\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \  -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4
-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9
-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}

\mathrm{Convert}\:x\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert}\:y\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \  -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1

Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1

For me I used
\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1
As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1

Hope this helps!
4 0
4 years ago
On a coordinate plane, a curved line with a minimum value of (negative 1.25, negative 3.25) and a maximum value of (0.25, negati
soldier1979 [14.2K]

The statement which is true about the end behavior of the graphed function is third option that is as the x-values go to negative infinity, the function's values are equal to zero.

A coordinate plane is a two dimensional plane formed by the intersection of two number lines . One of these number lines is a horizontal number lines called the x-axis and the other number line is a vertical number line called the y-axis.

The exact answer is third option that is as the x-values go to negative infinity, the function's values are equal to zero.

Step by step explanation:

Minimum value (negative 1.25, negative 3.25)

Maximum value of (0.25, negative 1.75)

The x intercepts where the graph crosses the x axis.

Value of x when y=0 crosses x axis at (negative 2.25, 0)

The y intercept are where the graph crosses the y axis.

Value of y when x=0 crosses y axis at (0, negative 2)

The line exits the plane at (negative 2.75, 6) and (1.5, 6).

It can be seen from the coordinates that as the x-values go to negative infinity, the function's values are equal to zero, the coordinates of y approaches zero.

Learn more about coordinates from:

brainly.com/question/17206319

#SPJ10

8 0
2 years ago
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