Any help i did not understand it can any body help me it will be great and thanks

Can someone please answer these questions?

Arisa [49]

Keep in mind that you can gather only like terms. Two terms can be summed/subtracted if they involve the same powers of the same variables.

So, going point by point:

We can subtract the terms, because they both involve the variable $b$ , and we have $3b-7b = (3-7)b = -4b$
We can sum the terms, because they all involve the variable $a$ , and we have $a+a-5a = (1+1-5)a = -3a$
We can't sum this terms: 3 is a pure number, while 3r involves the variable r.
We can only sum the two terms: we have $7t-8t-8 = (7-8)t-8 = -t-8$
We can't sum this terms: -20 is a pure number, while -15p involves the variable p.
We can sum the terms, because they all involve the variable $g$ , and we have $-g-g-g = (-1-1-1)g = -3g$

7 0

3 years ago

Previously, an organization reported that teenagers spent 24.5 hours per week, on average, on the phone. The organization thinks

Lena [83]

Answer:

We need to develop a one-tail t-student test ( test to the right )

We reject H₀ we find evidence that student spent more than 24,5 hours on the phone

Step-by-step explanation:

Sample size n = 15 n < 30

And we were asked if the mean is higher than, therefore is a one-tail t-student test ( test to the right )

Population mean μ₀ = 24,5

Sample mean μ = 25,7

Sample standard deviation s = 2

Hypothesis Test:

Null Hypothesis H₀ μ = μ₀

Alternative Hypothesis Hₐ μ > μ₀

t (c) = ?

We will define CI = 95 % then α = 5 % α = 0,05 α/2 = 0,025

n = 15 then degree of freedom df = 14

From t-student table we get: t(c) = 2,1448

And t(s)

t(s) = ( μ - μ₀ ) / s/√n

t(s) = (25,7 - 24,5) /2/√15

t(s) = 2,3237

Now we compare t(c) and t(s)

t(c) = 2,1448 t(s) = 2,3237

t(s) > t(c)

Then we are in the rejection region we reject H₀ we have evidence at 95% of CI that students spend more than 24,5 hours per week on the phone

8 0

3 years ago

The sum of the angle measures of a quadrilateral is 360 degree . Write and solve an equation to find the value of x . the two nu

trapecia [35]

Answer:

Step-by-step explanation:

I’m not too good with this but the answer should be 87.5

8 0

3 years ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

3 years ago

Find b. Question 19 options: 70.1 57.7 31.5 43.8

Reptile [31]

Answer: 57.7

Step-by-step explanation:

6 0

3 years ago