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3 years ago

Mrs. Jacobson asked her students to write a word problem that describes the graph below.

Mathematics

2 answers:

liraira [26]3 years ago

8 0

Of the options given, it's A) Jamie has $8, and she pays $2 per day to ride the bus.

Anna [14]3 years ago

6 0

The answer for the exercise shown above is the first option (Option A), which is: <span>A. Jamie has $8, and she pays $2 per day to ride the bus.
The explanation is shown below:
If you write the values given in the graph in a table, you can notice that the money decreases on each day. This means that the person does not earn money, but pays. On day zero (0), the person has $8 and then that person pays $2 per day. Therefore, the option A is the answer.</span>

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N/2 + 3 = 7 (n/2 is a fraction)

pishuonlain [190]

Answer:

n=8

Step-by-step explanation:

n/2+3=7

-3 -3

n/2 = 4

*2 *2

n=8

3 0

2 years ago

Sor-rey for posting again,I messed up on my last one...anyway,100 points and Brainlest! Please help ASAP!

Triss [41]

The answer of 4(2x^3 y^4 )^4/2(2x^2 y^6 )^3 after simplification will be

4 $x^{6} y^{-2}$

Step-by-step explanation:

To simplify

$\frac{4(2x^3 y^4 )^4}{2(2x^2 y^6 )^3}$ .

Using exponential law the exponents of variables will be added when multiplied as per rule

we get

$\frac{4[(2^4)(x^{3\times 4}) ( y^{4\times4} )] }{2[(2^3)(x^{2\times3} ) (y^{6\times3} )]}\\$

$\frac{4[(16)(x^{12}) ( y^{16} )] }{2[(8)(x^{6} ) (y^{18} )]}\\\\\frac{64x^{12}y^{16} }{16 x^{6}y^{18} }$

Now by dividing according to exponential rule the exponents of variables will be subtracted as per rule

$(\frac{64}{16} ) x^{12-6} y^{16-18} \\4 x^{6} y^{-2}$

So the answer of 4(2x^3 y^4 )^4/2(2x^2 y^6 )^3 after simplification will be

4 $x^{6} y^{-2}$

Keywords: Algebra

Learn more about algebra at:

#learnwithBrainly

6 0

2 years ago

Liz buys a sweater for $32. She pays 5% sales tax on the sweater.

pashok25 [27]

Answer:

$33.60 would be the total cost

6 0

2 years ago

Read 2 more answers

Use the continuous compound interest formula to find the indicated value. A=90,000; P=65,452; r=9.1%; t=? t=years (Do not round

Butoxors [25]

$\bf \qquad \textit{Simple Interest Earned}\\\\
A=Pe^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &90,000\\
P=\textit{original amount}\to& \$65,452\\
r=rate\to 9.1\%\to \frac{9.1}{100}\to &0.091\\
t=years
\end{cases}$

$\bf 90000=65452e^{0.091t}\implies \cfrac{90000}{65452}=e^{0.091t}\implies \cfrac{22500}{16363}=e^{0.091t}
\\\\\\
\textit{taking \underline{ln} to both sides}\qquad ln\left( \cfrac{22500}{16363} \right)=ln\left( e^{0.091t} \right)
\\\\\\
ln\left( \cfrac{22500}{16363} \right)=0.091t\implies 
\cfrac{ln\left( \frac{22500}{16363} \right)}{0.091}=t$

7 0

3 years ago

Subtract and simplify: (10x4 – 2x + 7) – (6x4 + 2x3 – 4x2 – 1) A. 4x4 – 2x3 + 4x2 – 2x + 6 B. 4x4 + 2x3 – 4x2 – 2x + 8 C. 4x4 –

Masja [62]

(10x^4 - 2x + 7) - (6x^4 + 2x^3 - 4x^2 - 1)=

$= 10x^4-2x+7-6x^4-2x^3+4x^2+1=\\\\ = 4x^4-2x^3+4x^2-2x+8$

8 0

2 years ago