There are 39 students in Ms. Salazar's chemistry class. If Ms. Salazar divides the class into 9 lab

one student wrote 18*d=54 for the sentence "The products of 18 and d equals 54". Another student wrote d*18=54 for the same sent

Aloiza [94]

Answer:

Both are right.

Step-by-step explanation:

The two are both right.

Multiplication in algebra obeys the cummutative law ( a * b = b * a).

5 0

3 years ago

Read 2 more answers

I need this asap ty:p

Anastasy [175]

Answer:

(a) x = -2y

(c) 3x - 2y = 0

Step-by-step explanation:

You can tell if an equation is a direct variation equation if it can be written in the format y = kx.

Note that there is no addition and subtraction in this equation.

Let's put these equations in the form y = kx.

(a) x = -2y

y = x/-2 → y = -1/2x
This is equivalent to multiplying x by -1/2, so this is an example of direct variation.

(b) x + 2y = 12

2y = 12 - x
y = 6 - 1/2x
This is not in the form y = kx since we are adding 6 to -1/2x. Therefore, this is <u>NOT</u> an example of direct variation.

(c) 3x - 2y = 0

-2y = -3x
y = 3/2x
This follows the format of y = kx, so it is an example of direct variation.

(d) 5x² + y = 0

y = -5x²
This is not in the form of y = kx, so it is <u>NOT</u> an example of direct variation.

(e) y = 0.3x + 1.6

1.6 is being added to 0.3x, so it is <u>NOT</u> an example of direct variation.

(f) y - 2 = x

y = x + 2
2 is being added to x, so it is <u>NOT</u> an example of direct variation.

The following equations are examples of direct variation:

x = -2y
3x - 2y = 0

7 0

3 years ago

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Hey can you help me answer this question by giving me the answer?

Leto [7]

The value of f(a)=4-2a+6 $a^{2}$ , f(a+h) is $6a^{2} +6h^{2} -2a-2h+12ah$ , [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6 $x^{2}$ .

Given a function f(x)=4-2x+6 $x^{2}$ .

We are told to find out the value of f(a), f(a+h) and [f(a+h)-f(a)]/hwhere h≠0.

Function is like a relationship between two or more variables expressed in equal to form.The value which we entered in the function is known as domain and the value which we get after entering the values is known as codomain or range.

f(a)=4-2a+6 $a^{2}$ (By just putting x=a).

f(a+h)== $4-2(a+h)+6(a+h)^{2}$

=4-2a-2h+6( $a^{2} +h^{2} +2ah$ )

=4-2a-2h+6 $a^{2} +6h^{2} +12ah$

= $6a^{2} +6h^{2}-2a-2h+12ah$

[f(a+h)-f(a)]/h=[ $6a^{2} +6h^{2}-2a-2h+12ah$ -(4-2a+6 $a^{2}$ )]/h

= $(6a^{2} +6h^{2} -2a-2h+12ah)/h$

= $(6h^{2} -2h+12ah)/h$

=6h+12a-2.

Hence the value of function f(a)=4-2a+6 $a^{2}$ , f(a+h) is $6a^{2} +6h^{2} -2a-2h+12ah$ , [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6 $x^{2}$ .

Learn more about function at brainly.com/question/10439235

#SPJ1

7 0

2 years ago

Lily is a botanist who works for a garden that many tourists visit. The function f(s) = 2s + 30 represents the number of flowers

Vesnalui [34]

Answer:

A. b(w) = 80w +30

B. input: weeks; output: flowers that bloomed

C. 2830

Step-by-step explanation:

For f(s) = 2s +30, and s(w) = 40w, the composite function f(s(w)) is ...

b(w) = f(s(w)) = 2(40w) +30

b(w) = 80w +30 . . . . . . blooms over w weeks

__

The input units of f(s) are <em>seeds</em>. The output units are <em>flowers</em>.

The input units of s(w) are <em>weeks</em>. The output units are <em>seeds</em>.

Then the function b(w) above has input units of <em>weeks</em>, and output units of <em>flowers</em> (blooms).

__

For 35 weeks, the number of flowers that bloomed is ...

b(35) = 80(35) +30 = 2830 . . . . flowers bloomed over 35 weeks

7 0

2 years ago

NEED HELP ASAP!!!!!!!!!!!!!!!!!!!!!!

Georgia [21]

The third table on the top row. It has a linear function because the Y is always constant. For example: -3 + -2= -5 -5 + -2= -7

7 0

3 years ago