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3 years ago

NEW ones help me pls

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

3 0

Answer:

the aswer is D

Step-by-step explanation:

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HELP QUICK!!!!!!!!!!

Mama L [17]

Answer:

Option is right

Step-by-step explanation:

Given that bella want to know what the students at her school think of her show. which is the BEST group to survey to find this information

For getting accurate results she must select a random sample without bias

For this option a will not serve since she selects only students she recognizes.

Option B may not be right, as during show on the face they may not give correct opinion

C is right as a random sample is taken

D is wrong as the first 30 students may not be at random.

So option C is right

8 0

3 years ago

Read 2 more answers

Which statement correctly compares the two functions? GIVING BRAINLIEST!!

geniusboy [140]

Answer:

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

How do you solve this?

nalin [4]

So u have to add all the sides up.
which will give you 6 + 6√3 + 6 +6√3
because their opposite and parallel lengths and widths are equal
that gives you 12 + 12√3
so ans is 12m + 12√3m

7 0

2 years ago

I WILL GIVE BRAINLEST!!! (6 points!)

vlada-n [284]

Answer:

OPTION A: 2x + 3y = 5

Step-by-step explanation:

The product of slopes of two perpendicular lines is -1.

We rewrite the given equation as follows:

2y = 3x + 2

⇒ y = $\frac{3}{2}x + 1$

The general equation of the line is: y = mx + c, where 'm' is the slope of the line.

Here, m = $\frac{3}{2}$ .

Therefore, the slope of the line perpendicular to the line given = $\frac{-2}{3}$ because $\frac{3}{2} \times \frac{-2}{3} = -1$ .

To determine the equation of the line passing through the given point and a slope we use the Slope - One - point formula which is:

y - y₁ = m(x - x₁)

The point is: (x₁, y₁) = (-2, 3)

Therefore, the equation is:

y - 3 = $\frac{-2}{3}$ (x + 2) $

⇒ 3y - 9 = -2(x + 2)

⇒ 3y - 9 = -2x - 4

⇒ 2x + 3y = 5 is the required equation.

6 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B

inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$

$\rule{190pt}{2pt}$

5 0

2 years ago