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vredina [299]
4 years ago
9

I need help with this

Mathematics
1 answer:
Assoli18 [71]4 years ago
7 0
On a given line, (on one side) there are a total of 180°

if one line in Problem #3 is bisected by a line, with one half X and the other 120°,

do 180° (the total) minus 120° which=60°

now the hard part, that line that bisected the first line is bisecting a line that is parallel to your second line, the one with <5 and <6

this means that the big angle formed in the first one with 120° is the same angle as in the second line, leaving <5 as 120°
which means <6 is 60°, like in the top part of the problem. You're basically flipping the top line upside down, I hope it helps.
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4-5 of the money collected from a fundraiser was dived equally among 8 grades.What fraction of money did each grade receive
allsm [11]

If you are just dividing percent's, then each grade would have 12.5 percent.

4 0
4 years ago
Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1
sleet_krkn [62]

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}

From Left side:

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)

\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}

NOTE: sin²θ + cos²θ = 1

\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{2}{sin^2\theta-cos^2\theta}

\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}

\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}

\dfrac{2sec^2\theta}{tan^2\theta-1}

Left side = Right side <em>so proof is complete</em>

8 0
3 years ago
Read 2 more answers
What is the rate of decay for the function f(x) =2(0.95)^x ( x is on top of the parentheses)
SIZIF [17.4K]

Answer:

The decay rate is 5%.

Step-by-step explanation:

Let a substance is decaying at the rate of r% per hour from the initial value of P for t hours, then the final value of the substance is given by the function  

f(t) = P(1 - \frac{r}{100})^{t} ........... (1)

Comparing this equation with the original equation given as

f(x) = 2(0.95)^{x} ............ (2) we get,

(1 - \frac{r}{100}) = 0.95

⇒ \frac{r}{100} = 0.05

⇒ r = 5%.

Therefore, the decay rate is 5%. (Answer)

7 0
3 years ago
Let f(x)=6(2)x−1+4
atroni [7]
G(x)=6(2)x+7+4 i think that is it

5 0
3 years ago
What is the solution set for the given inequality if the replacement set for r is {5, 6, 7, 8, 9, 10}? 2r ≤ 3r – 8 A. {8, 9, 10}
TiliK225 [7]
2r\leq3r-8\\&#10;r\geq8\\\Downarrow\\&#10;\boxed{\{8,9,10\}}
8 0
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