Please answer ASAP!! Thank you!!:)

Mathematics

2 answers:

prisoha [69]3 years ago

8 0

Answer: T -AXIS = 440

H-AXIS = 40

Step-by-step explanation:

Hi, to solve this we have to analyze the information given:

Hours worked per week: between 10 and 40 hours

He is paid $11 an hour

So, the scale for the T-axis (total pay) of his graph should go from 0 to at least 440. We obtain this number by multiplying $11 (the price of an hour) by the maximum of hours he can work in a week (40).

Mathematically speaking: $11 x 40 = 440

And for the h-axis of his graph, it should go from 0 to at least 40, which is the maximum number of hours he can work in a week.

aksik [14]3 years ago

5 0

the top of his graph for T should be at least

11 * 40 = 440

the h axis should be at least the number of hour he works

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Find the value of the following expression: (2^8 ⋅ 5^−5 ⋅ 19^0)^−2 ⋅ 5 to the power of negative 2 over 2 to the power of 3, whol

koban [17]

Answer:

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

Step-by-step explanation:

$\left(2^8\cdot5^{-5}\cdot19^0\right)^{-2}\cdot\left(\dfrac{5^{-2}}{2^3}\right)^4\cdot228\\\\\text{use}\ a^{-n}=\dfrac{1}{a^n}\ \text{and}\ a^0=1\ \text{and}\ (a^n)^m=a^{nm}\\\\=\left(2^8\cdot\dfrac{1}{5^5}\cdot1\right)^{-2}\cdot\left(\dfrac{\frac{1}{5^2}}{2^3}\right)^4\cdot228=\left(\dfrac{2^8}{5^5}\right)^{-2}\cdot\left(\dfrac{1}{2^35^2}\right)^4\cdot228$

$=\dfrac{(2^8)^{-2}}{(5^5)^{-2}}\cdot\dfrac{1^4}{(2^3)^4(5^2)^4}\cdot228=\dfrac{2^{-16}}{5^{-10}}\cdot\dfrac{1}{2^{12}5^8}\cdot228\\\\\text{use}\ a^n=\dfrac{1}{a^{-n}}\to\dfrac{1}{a^n}=a^{-n}\\\\=2^{-16}\cdot5^{10}\cdot2^{-12}\cdot5^{-8}\cdot228\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{-16+(-12)}\cdot5^{10+(-8)}\cdot228=2^{-28}\cdot5^2\cdot228\\\\=2^{-28}\cdot5^2\cdot4\cdot57=2^{-28}\cdot5^2\cdot2^2\cdot57=2^{-28+2}\cdot5^2\cdot57\\\\=2^{-26}\cdot5^2\cdot57=\dfrac{5^2\cdot57}{2^{26}}$

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

4 0

3 years ago

The function f(x) = –x2 + 20x – 75 models the profit from one customer, in dollars, a shop makes for printing photos, where x is

trasher [3.6K]

The function is graphed as shown below

Part A:

We use the formula $- \frac{b}{2a}$ to find the vertex of the function. A quadratic function of the form of $a x^{2} +bx+c$ and equating this form to the given function $- x^{2} +20x-75$ , we have
$b=20$ and $a=-1$ .

Substituting $a$ and $b$ into the vertex formula, we have
$x=- \frac{20}{2(-1)}=10$ , as shown in the graph

This calculation means that the highest profit is achieved when the number of photo printed equals to ten photos

Part B:

We can find solution to this equation by factorising

$- x^{2} +20x-75=0$
$-( x^{2} -20x+75)=0$
$x^{2} -20x+75=0$
$(x-5)(x-15)=0$
$x=5$ and $x=15$ , as shown in the graph

The two values means that the company makes no profit when they either produce 5 or 15 photos