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mina [271]
3 years ago
5

Please help me with this geometry question image attached

Mathematics
1 answer:
iren [92.7K]3 years ago
6 0

For this problem, you need to know how to use sin, cos, and tan or 'SOH-CAH-TOA' sin- opposite/hypotenuse cos-adjacent/hypotenuse toa- opposite/adjacent

in this case, you can use tan or toa

the equation should be set up like this:

4/x=1/0.848

solve, and you will get 3.39...

(you need to use the sin,cos,tan degree chart to find 0.848)

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olchik [2.2K]
The reading is 26°c as the small sections r of difference 2
5 0
3 years ago
\A and \B are supplementary angles. If m\A= (3x – 23) and m
mrs_skeptik [129]

2 supplementary angles when added together need to equal 180 degrees.

Add:

3x-23 + 2x-12 = 180

Simplify:

5x -35 = 180

Add 35 to both sides:

5x = 215

Divide both sides by 5:

X = 43

Now replace x with 43 in the equation for angle b:

2(43) -12 = 86-12 = 74 degrees

Angle B = 74 degrees

5 0
3 years ago
Which number is divisible by both three and nine?
horrorfan [7]

Answer:

765 is divisible by 3 and 9 I hope that this helps

Step-by-step explanation:

765÷9=85

765÷3=255

7 0
2 years ago
Solve y=-3x² + 4.5x – 20<br> Use the Quadratic formula
statuscvo [17]

Answer:

x=0.75 \pm 2.47067i

Step-by-step explanation:

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

√-1 is imaginary number i

Step 1: Define

y = -3x² + 4.5x - 20

a = -3

b = 4.5

c = -20

Step 2: Substitute and Evaluate

x=\frac{-4.5 \pm \sqrt{4.5^2-4(-3)(-20)} }{2(-3)}

x=\frac{-4.5 \pm \sqrt{20.25-240} }{-6}

x=\frac{-4.5 \pm \sqrt{-219.75} }{-6}

x=\frac{-4.5 \pm \sqrt{219.75}(\sqrt{-1} ) }{-6}

x=\frac{-4.5 \pm i\sqrt{219.75} }{-6}

x=\frac{4.5 \pm i14.824 }{6}

x=0.75 \pm 2.47067i

7 0
3 years ago
Find the exact length of the curve. 36y2 = (x2 − 4)3, 5 ≤ x ≤ 9, y ≥ 0
IrinaK [193]
We are looking for the length of a curve, also known as the arc length. Before we get to the formula for arc length, it would help if we re-wrote the equation in y = form.

We are given: 36 y^{2} =( x^{2} -4)^3
We divide by 36 and take the root of both sides to obtain: y = \sqrt{ \frac{( x^{2} -4)^3}{36} }

Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain: y =  \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}

Let's leave that for the moment and look at the formula for arc length. The formula is L= \int\limits^c_d {ds} where ds is defined differently for equations in rectangular form (which is what we have), polar form or parametric form.

Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows: ds= \sqrt{1+( \frac{dy}{dx})^2 } dx

As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.

As you can see from the formula we need to find dy/dx and square it. Let's do that now.

We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here x^2-4). More formally, we can let u=x^{2} -4 and then consider the derivative of u^{3/2}du. Either way, we obtain,

\frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2}

Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4}

This means that in our case:
ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{x^4-4 x^{2}+4 }{4}} dx
ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx
ds=  \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx

Recall, the formula for arc length: L= \int\limits^c_d {ds}
Here, the limits of integration are given by 5 and 9 from the initial problem (the values of x over which we are computing the length of the curve). Putting it all together we have:

L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x evaluated from 9 to 5 (I cannot seem to get the notation here but usually it is a straight line with the 9 up top and the 5 on the bottom -- just like the integral with the 9 and 5 but a straight line instead). This means we plug 9 into the expression and from that subtract what we get when we plug 5 into the expression.

That is, [(\frac{1}{2}) ( \frac{9^3}{3}) -9]-([(\frac{1}{2}) ( \frac{5^3}{3}) -5]=( \frac{9^3}{6}-9)-( \frac{5^3}{6}-5})=\frac{290}{3}


8 0
4 years ago
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