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3 years ago

What function would this be ?

Mathematics

2 answers:

olga2289 [7]3 years ago

8 0

I believe it would be D. a = 100 - 16d.

Because you have 100 dollars overall and you spend 16 dollars per day.

If you plug 4 (for the number of days that 16 dollars was spent) into the equation, you will find out if you have enough money left. Let's see!

a = 100 - 16d

a = 100 - 16(4)

a = 100 - 64

a = 36

Yay! You have exactly 36 dollars left! You can buy the sweatshirt!

Hope this helps!

Vadim26 [7]3 years ago

4 0

Is letter D because is saying that spending 16 each day so is minus 16

You might be interested in

Melissa decided to divide her $42,000 of savings into three investments: a savings account paying 5% simple interest, a time dep

Marat540 [252]

Answer:

Amount in savings account = $24,000

amount put into time deposit account = $11,000

amount put into bond account = $7,000

Step-by-step explanation:

Let x be amount put into savings account, y be amount put into time deposit account and z be amount put into bond account.

Thus;

0.05x + 0.07y + 0.09z = 2600 - - (eq1)

Also,

x + y + z = 42,000 - - - (eq2)

Now,we are told she invested 6000 more in the savings account than the other 2 combined. Thus;

x = 6000 + y + z

So,

x - y - z = 6000 - - - (eq3)

Solving the 3 equations simultaneously gives;

x = $24,000 ; y = $11,000 ; z = $7,000

5 0

4 years ago

Lincoln rode his bike 16 miles in 104 minutes. If he rode at a constant speed how far did he ride in 10 minutes

Olegator [25]

Use cross multiplication to solve:

$\frac{104}{10} =\frac{16}{x}\\ \\10*16=160\\104x=160\\\\x=160/104\\x=\frac{20}{13}$

So, he rode $\frac{20}{13}$ miles in 10 minutes (approximately 1.54mi).

3 0

4 years ago

WILL MARK BRAINLIEST IF CORRECT!

ziro4ka [17]

Answer:

Answer → Student A was right.

${ \tt{ \sqrt[3]{1944} = \sqrt[3]{(72 \times 27)} }} \\ \\ = { \tt{ \sqrt[3]{72} \times \sqrt[3]{27} }} \\ \\ = { \tt{ \sqrt[3]{72} \times 3}} \\ \\ = { \underline{ \tt{ \: 3 \sqrt[3]{72} } \: }}$

8 0

3 years ago

Read 2 more answers

Could someone help me? the video doenst explain it

True [87]

Personally, I would recommend reading this as a fraction, and splitting up mantissa from the power of 10. (Mantissa refers to the number 1.23 in $1.23\times10^n$ .)

$\left(3.6 \times 10^{-5}\right) \div \left(9 \times 10^{-7}\right) = \dfrac{3.6 \times 10^{-5}}{9 \times 10^{-7}} = \dfrac{3.6}9 \times \dfrac{10^{-5}}{10^{-7}}$

Now just reduce each fraction:

$\dfrac{3.6}9 = \dfrac{36}{90} = \dfrac25 = 0.4$

$\dfrac{10^{-5}}{10^{-7}} = 10^{-5-(-7)} = 10^{-5+7} = 10^2$

$\left(3.6 \times 10^{-5}\right) \div \left(9 \times 10^{-7}\right) = 0.4 \times 10^2 = \boxed{4 \times 10^1}$

since $0.4\times10^2 = 0.4 \times 100 = 40 = 4 \times 10$ .

7 0

3 years ago

This 1 seems really complicated

Fofino [41]

The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
________________________________________________________
Given:
________________________________________________________
y = - 4x + 16 ;

4y − x + 4 = 0 ;
________________________________________________________
"Solve the system using substitution" .
________________________________________________________
First, let us simplify the second equation given, to get rid of the "0" ;

→ 4y − x + 4 = 0 ;

Subtract "4" from each side of the equation ;

→ 4y − x + 4 − 4 = 0 − 4 ;

→ 4y − x = -4 ;
________________________________________________________
So, we can now rewrite the two (2) equations in the given system:
________________________________________________________

y = - 4x + 16 ; ===> Refer to this as "Equation 1" ;

4y − x = -4 ; ===> Refer to this as "Equation 2" ;
________________________________________________________
Solve for "x" and "y" ; using "substitution" :
________________________________________________________
We are given, as "Equation 1" ;

→ " y = - 4x + 16 " ;
_______________________________________________________
→ Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;

to solve for "x" ; as follows:
_______________________________________________________
Note: "Equation 2" :

→ " 4y − x = - 4 " ;
_________________________________________________
Substitute the value for "y" {i.e., the value provided for "y"; in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;

→ as follows:
_________________________________________________

→ " 4 (-4x + 16) − x = -4 " ;
_________________________________________________
Note the "distributive property" of multiplication :
_________________________________________________

a(b + c) = ab + ac ; AND:

a(b − c) = ab − ac .
_________________________________________________
As such:

We have:

→ " 4 (-4x + 16) − x = - 4 " ;
_________________________________________________
AND:

→ "4 (-4x + 16) " = (4* -4x) + (4 *16) = " -16x + 64 " ;
_________________________________________________
Now, we can write the entire equation:

→ " -16x + 64 − x = - 4 " ;

Note: " - 16x − x = -16x − 1x = -17x " ;

→ " -17x + 64 = - 4 " ; Solve for "x" ;

Subtract "64" from EACH SIDE of the equation:

→ " -17x + 64 − 64 = - 4 − 64 " ;

to get:

→ " -17x = -68 " ;

Divide EACH side of the equation by "-17" ;
to isolate "x" on one side of the equation; & to solve for "x" ;

→ -17x / -17 = -68/ -17 ;

to get:

→ x = 4 ;
______________________________________
Now, Plug this value for "x" ; into "{Equation 1"} ;

which is: " y = -4x + 16" ; to solve for "y".
______________________________________

→ y = -4(4) + 16 ;

= -16 + 16 ;

→ y = 0 .
_________________________________________________________
The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
_________________________________________________________
Now, let us check our answers—as directed in this very question itself ;
_________________________________________________________
→ Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten;

→ Let us check;

→ For EACH of these 2 (TWO) equations; do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ;

→ Consider the first equation given in our problem, as originally written in the system of equations:

→ " y = - 4x + 16 " ;

→ Substitute: "4" for "x" and "0" for "y" ; When done, are both sides equal?

→ "0 = ? -4(4) + 16 " ?? ; → "0 = ? -16 + 16 ?? " ; → Yes! ;

{Actually, that is how we obtained our value for "y" initially.}.

→ Now, let us check the other equation given—as originally written in this very question:

→ " 4y − x + 4 = ?? 0 ??? " ;

→ Let us "plug in" our obtained values into the equation;

{that is: "4" for the "x-value" ; & "0" for the "y-value" ;

→ to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.

→ " 4(0) − 4 + 4 = ? 0 ?? " ;

→ " 0 − 4 + 4 = ? 0 ?? " ;

→ " - 4 + 4 = ? 0 ?? " ; Yes!
_____________________________________________________
→ As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
_____________________________________________________
→ "x = 4" and "y = 0" ; or; write as: [0, 4] ; are correct.
_____________________________________________________
Hope this lenghty explanation is of help! Best wishes!
_____________________________________________________

7 0

3 years ago