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riadik2000 [5.3K]

2 years ago

7

Plz help ASAP!!!!!!!!!!! IT HAS TO BE CORRECT!

1 answer:

iren [92.7K]2 years ago

3 0

Answer:

(9,8)

Step-by-step explanation:

multiply both x and y by -1 and add 5 to the y value

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you want to put 520 quarters in coin wrappers. you need one wrapper for every $10 in quarters. write an equation you can use to

Lisa [10]

40c=520
This is the proper equation because there are 40 quarters in $10. Thus, if we could only fit $10 worth of quarters in each wrapper, we can only fit 40quarters in each wrapper. C represents the amount of wrappers we will need.

Hope this helps!

6 0

2 years ago

The sum of twice a number and 6 is equal to the sum of the number and 3

kvasek [131]

2x+6=x+3 this is the answer

8 0

2 years ago

Probe that:<br><img src="https://tex.z-dn.net/?f=%20%5Csec%20%5Calpha%20%20%5Csqrt%7B1%20-%20%20%5Csin%28%20%7B%7D%5E%7B2%7D%20%

Nataly [62]

Step-by-step explanation:

<h3> $\sec \alpha \sqrt{1 - \sin ^{2} \alpha } = 1$ </h3>

Prove the LHS

Using trigonometric identities

That's

<h3> $\cos ^{2} \alpha = 1 - \sin^{2} \alpha$ </h3>

<u>Rewrite the expression</u>

We have

<h3> $\sec \alpha \sqrt{ \cos^{2} \alpha }$ </h3>

<h3> $\sqrt{ { \cos }^{2} \alpha } = \cos \alpha$ </h3>

So we have

<h3> $\sec \alpha \times \cos \alpha$ </h3>

Using trigonometric identities

<h3> $\sec \alpha = \frac{1}{ \cos \alpha }$ </h3>

<u>Rewrite the expression</u>

That's

<h3> $\frac{1}{\cos \alpha } \times \cos \alpha$ </h3>

Reduce the expression with cos a

We have the final answer as

<h2>1</h2>

As proven

Hope this helps you

7 0

2 years ago

Please help! Don't put anything random for the points I really need help!

Kamila [148]

Answer:

x = -2 and x = 5/4

Step-by-step explanation:

The image below shows step-by-step on how to solve it.

Hope this helps! :)

8 0

2 years ago

What are the vertical asymptotes of f(x) = 10/ x^2 - 1

Veronika [31]

Vertical asymptotes ares found where the function is undefined. f(x) is undefined when the denominator equals zero:

x^2-1=0

<em>*Add 1 to both sides*</em>

x^2=1

<em>*Take the square root of both sides*</em>

x=+/-1

There are vertical asymptotes at x=-1 and x=1.

Hope this helps!!

8 0

3 years ago