Please I need help do not understand the 3,4,5,6

What’s the solution of x+y=29 and y=4x-11

tia_tia [17]

Answer:

x = 8

y = 21

Step-by-step explanation:

y = 4x − 11

x + y = 29

Solve for x:

x + y = 29

x + 4x − 11 = 29

5x − 11 = 29

5x = 40

x = 8

Solve for y:

y = 4x − 11

y = 4 × 8 − 11

y = 21

6 0

3 years ago

[Pre-Calc] Please Help! I don’t know where to start. How do I do this?

sertanlavr [38]

Answer:

See Below.

Step-by-step explanation:

Problem A)

We have:

$\displaystyle \csc^2\theta \tan^2\theta -1=\tan^2\theta$

When in doubt, convert all reciprocal trig functions and tangent into terms of sine and cosine.

So, let cscθ = 1/sinθ and tanθ = sinθ/cosθ. Hence:

$\displaystyle \left(\frac{1}{\sin^2\theta}\right)\left(\frac{\sin^2\theta}{\cos^2\theta}\right)-1=\tan^2\theta$

Cancel:

$\displaystyle \frac{1}{\cos^2\theta}-1=\tan^2\theta$

Let 1/cosθ = secθ:

$\sec^2\theta -1=\tan^2\theta$

From the Pythagorean Identity, we know that tan²θ + 1 = sec²θ. Hence, sec²θ - 1 = tan²θ:

$\tan^2\theta =\tan^2\theta$

Problem B)

We have:

$\sin^3x=\sin x-\sin x \cos^2 x$

Factor out a sine:

$\sin x(\sin^2 x)=\sin x-\sin x\cos^2 x$

From the Pythagorean Identity, sin²θ + cos²θ = 1. Hence, sin²θ = 1 - cos²θ:

$\sin x(1-\cos^2 x)=\sin x-\sin x\cos^2x$

Distribute:

$\sin x- \sin x \cos^2 x=\sin x-\sin x\cos^2 x$

Problem C)

We have:

$\displaystyle \frac{\cos 2x+1}{\sin 2x}=\cot x$

Recall that cos2θ = cos²θ - sin²θ and that sin2θ = 2sinθcosθ. Hence:

$\displaystyle \frac{\cos^2 x-\sin^2 x+1}{2\sin x\cos x}=\cot x$

From the Pythagorean Identity, sin²θ + cos²θ = 1 so cos²θ = 1 - sin²θ:

$\displaystyle \frac{2\cos^2 x}{2\sin x\cos x}=\cot x$

Cancel:

$\displaystyle \frac{\cos x}{\sin x}=\cot x$

By definition:

$\cot x = \cot x$

3 0

3 years ago

Please help me.

creativ13 [48]

1.is A
2. is D
3. is A
4. is B
i checked these answers

6 0

3 years ago

Read 2 more answers

Hurry please its timed<br> Write a biconditional statement that is true.

slega [8]

Answer:

If a polygon has three sides, then it must be a triangle.

Step-by-step explanation:

use if-then statements to make it biconditional

4 0

3 years ago

Given the functions f and g, explain why you cannot find f⋅g.

Delvig [45]

We can't write the product because there is no common input in the tables of g(x) and f(x).

<h3>Why you cannot find the product between the two functions?</h3>

If two functions f(x) and g(x) are known, then the product between the functions is straightforward.

g(x)*f(x)

Now, if we only have some coordinate pairs belonging to the function, we only can write the product if we have two coordinate pairs with the same input.

For example, if we know that (a, b) belongs to f(x) and (a, c) belongs to g(x), then we can get the product evaluated in a as:

(g*f)(a) = f(a)*g(a) = b*c

Particularly, in this case, we can see that there is no common input in the two tables, then we can't write the product of the two functions.

If you want to learn more about product between functions:

brainly.com/question/4854699

#SPJ1

6 0

2 years ago

Read 2 more answers