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gtnhenbr [62]
3 years ago
14

Geometry help ASAP 15 points Need help to find A and B

Mathematics
1 answer:
Marta_Voda [28]3 years ago
6 0
sin60^o= \frac{a}{2 \sqrt{2} }  \ \ \to  \ a=2 \sqrt{2} *sin60^o=2 \sqrt{2}* \frac{ \sqrt{3}}{2}= \sqrt{6}     \\  \\ cos60^o =  \frac{b}{2 \sqrt{2} }  \ \ \to  \ b=2 \sqrt{2} *cos60^o=2 \sqrt{2}*0.5= \sqrt{2}

Answer: a=√6,  b=√2
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lisov135 [29]

The function is given f(x)=2^x+3.

Now let's solve for f(-2).

f(-2)=2^{-2}+3

Rule of negative exponent: x^{-1}=\frac{1}{x^1}.

f(-2)=\frac{1}{2^2}+3

f(-2)=\frac{1}{4}+\frac{3}{1}

Rule for sum of fractions: \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}.

f(-2)=\frac{13}{4}

And the result is:

f(-2)=\boxed{\frac{13}{4}=3.25}

4 0
3 years ago
17% of a number is 135
mel-nik [20]
794 is the number :)
3 0
3 years ago
Read 2 more answers
HELP PLEASE with this question
grin007 [14]
No it's not a function pretty sure
7 0
2 years ago
What is the property of 7+(a+b)=7+a)+b
-BARSIC- [3]

The <u>associative property of addition</u> says that when every operation is addition, you can group numbers however you like and choose which pair of numbers to add first; you can move parentheses without changing the answer.

7 0
3 years ago
Read 2 more answers
Find the equation of a line tangent to the circle x2+(y−3)2=34 at the point (5, 0).
salantis [7]

Answer:

Step-by-step explanation:

In order to find the equation of the line tangent to that circle, we have to find the derivative by implicit differentiation which will give us the slope formula of that tangent line.  Let's begin by expanding through the parenthesis to get the standard form of the circle:

x^2+y^2-6y+9=34

Moving the 9 to the other side since the derivative of a constant is 0 gives us

x^2+y^2-6y=25

By implicit differentation, the derivative of that function is

2x+2y\frac{dy}{dx}-6\frac{dy}{dx}=0

To find the derivative, we have to solve for dy/dx:

2y\frac{dy}{dx}-6\frac{dy}{dx}=-2x

Factor out the dy/dx to get:

\frac{dy}{dx}(2y-6)=-2x

Now divide to get your slope formula (first derivative):

\frac{dy}{dx}=\frac{-2x}{2y-6}

Now we can sub in the x and y values from the coodinate to get the slope of that tangent line:

\frac{dy}{dx}=\frac{-2(5)}{2(0)-6}=\frac{-10}{-6}=\frac{5}{3}

So now that have the slope, we can use the point-slope form of a line to write the equation of the tangent line.  The point-slop form of a line is:

y-y₁ = m(x-x₁)

Filling in we get:

y - 0 = 5/3(x - 5) so the equation of the tangent line is:

y=\frac{5}{3}x-\frac{25}{3}

Good luck in your calculus class!

3 0
3 years ago
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