All categories

4 years ago

What is the value of the function when x = 4? y = 3x + 7

2 answers:

8 0

2x - 3y = 7 and -3x + y = 7..multiply Equation 2 by THREE and add to Equation 1
-9x + 3y = 21...........................watch the y's disappear
-7x........ = 28
x = -4
substitute -4 instead of x in either of the ORIGINAL equations

2x - 3y = 7
2(-4) - 3y = 7
-8 -3y = 7..........add 8 to both sides
-3y = 15
y = -5

im not sure

denpristay [2]4 years ago

8 0

$y=3x+7\ \ \ | substitute\ x=4\\\\
y=3*4+7\\\\
y=12+7=19\\\\
Value\ of\ function\ is\ 19.$

You might be interested in

252 calories in 4 servings is how many calories per serving?

Lelu [443]

Answer:

63 calories

Step-by-step explanation:

Divide the number of calories by the number of servings

calories/servings

252/4

So, there are 63 calories per serving

Hope this helps!

8 0

3 years ago

Read 2 more answers

Why does a logarithmic equation sometimes have an extraneous solution?

jonny [76]

The main reason behind this is using properties of logarithm .

when solving

log_2(x+2)+log_2(x-3)=3

There you use multiplication property and make addition to multiplication then you get extraneous solution because in plenty of cases they occur

(-2)²=(2)²

But

-2≠2

Same happens in case of logarithm

6 0

2 years ago

Read 2 more answers

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago

15 POINTS!!!!

olganol [36]

p/5 + 6 = 10

=> p/5 = 10-6

=> p/5 = 4

=> p = 5×4

=> p = 20

C. 20

4 0

3 years ago

Read 2 more answers

Help me please i really need it!

Mademuasel [1]

Buddy what is the problem lol

6 0

3 years ago