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3 years ago

-3x + y = 1. written in genreal form how do i so this and what is the answer?!?!?!?!?

1 answer:

6 0

-3x + y = 1 ............. the general form is y= mx +b .................y = 1 + 3x ........................... Y = 3x + 1

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a tray is fill with several 500g bags of apples. the tray holds 8 kg in total. 5 bags of apples are sold .how many is left on th

kobusy [5.1K]

8kg=8000g

8000/500=16
16-2=14

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3 years ago

Write an addition sentence that helps you solve 15-9.

valentinak56 [21]

Answer:

9+6=15

9+1+5=15

Step-by-step explanation:

6 0

2 years ago

Jason and David went hiking at different times. Jason is currently 54 feet above sea level while David is starting at 11 feet be

tatuchka [14]

Since jason is currently at 54 feet above sea level and David is 11 feet below sea level the add 11 plus 54 which would give you 65 feet is there distance of elevations

3 0

3 years ago

Complete the table of values

Agata [3.3K]

Both problems give you a function in the second column and the x-values. To find out the values of a through f, you need to plug in those x-values into the function and simplify!

You need to know three exponent rules to simplify these expressions:
1) The negative exponent rule says that when a base has a negative exponent, flip the base onto the other side of the fraction to make it into a positive exponent. For example, $3^{-2} =
\frac{1}{3^{2} }$ .
2) Raising a fraction to a power is the same as separately raising the numerator and denominator to that power. For example, $(\frac{3}{4}) ^{3} = \frac{ 3^{3} }{4^{3} }$ .
3) The zero exponent rule says that any number raised to zero is 1. For example, $3^{0} = 1$ .


Back to the Problem:
Problem 1
The x-values are in the left column. The title of the right column tells you that the function is $y = 4^{-x}$ . The x-values are:
1) x = 0
Plug this into $y = 4^{-x}$ to find letter a:
$y = 4^{-x}\\
y = 4^{-0}\\
y = 4^{0}\\
y = 1$

2) x = 2
Plug this into $y = 4^{-x}$ to find letter b:
$y = 4^{-x}\\ 
y = 4^{-2}\\ 
y = \frac{1}{4^{2}} \\ 
y= \frac{1}{16}$

3) x = 4
Plug this into $y = 4^{-x}$ to find letter c:
$y = 4^{-x}\\ 
y = 4^{-4}\\ 
y = \frac{1}{4^{4}} \\ 
y= \frac{1}{256}$


Problem 2
The x-values are in the left column. The title of the right column tells you that the function is $y = (\frac{2}{3})^x$ . The x-values are:
1) x = 0
Plug this into $y = (\frac{2}{3})^x$ to find letter d:
$y = (\frac{2}{3})^x\\
y = (\frac{2}{3})^0\\
y = 1$

2) x = 2
Plug this into $y = (\frac{2}{3})^x$ to find letter e:
$y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^2\\ y = \frac{2^2}{3^2}\\
y = \frac{4}{9}$

3) x = 4
Plug this into $y = (\frac{2}{3})^x$ to find letter f:
$y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^4\\ y = \frac{2^4}{3^4}\\ y = \frac{16}{81}$

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Answers:
a = 1
b = $\frac{1}{16}$ 
c = $\frac{1}{256}$
d = 1
e = $\frac{4}{9}$
f = $\frac{16}{81}$

5 0

3 years ago

I Need Help With This One

Arisa [49]

Answer:

x = -6 x= -3

Step-by-step explanation:

f(x) = x^2+9x+18

Set equal to 0 to find the zeros

0 = x^2+9x+18

Factor, what numbers multiply to 18 and add to 9

6*3 =18

6+3=9

0= (x+6) (x+3)

Using the zero product property

x+6 =0 x+3 =0

x = -6 x= -3

7 0

2 years ago