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3 years ago

Please help with the question attached, will mark brainliest!!!!!!!!!!!

Mathematics

1 answer:

UkoKoshka [18]3 years ago

7 0

12; 4

66; 33.4

32.6

respectively in the blanks

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Help me if this is always true, sometimes true or never true pls<br> math

Angelina_Jolie [31]

23.
a. Sometimes true
b. Always true
c. Always true
d. Never true

Hope this helps. Good luck! :)

6 0

3 years ago

I need a equation for my parabola! I tried so many times and got it WroNg-

rodikova [14]

Answer:

Step-by-step explanation:

Not QUITE a parabola . but up until y = 4 it is close

Vertex = (-7-10)

Vertex form of parabola

y = a ( x+7)^2 -10 find a integer-coordinat point to calculate 'a'

I'll use -18,0

0 = a(-18+7)^2 -10

0 = 122 a -10 a = 10/122

y = 10/122 (x+7)^2 - 10 see graph below:

5 0

2 years ago

The number /3 goes on forever with no repeating pattern;therefore,it is rational

gizmo_the_mogwai [7]

Answer:

Actually this is irrational

Step-by-step explanation:

Irrational numbers continue forever without repeating for a good example would be pi

8 0

3 years ago

What's the value of x with 125° and 31°

astra-53 [7]

24.8% because 31 is the percentage of 125

6 0

3 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

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