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gregori [183]
3 years ago
10

The cashier counts 80 bills, all with the same value. She gets a total of $800. What is the value of each bill?

Mathematics
2 answers:
ololo11 [35]3 years ago
7 0

Answer:

10 dollars

Step-by-step explanation:

a_sh-v [17]3 years ago
5 0

Answer:

10!

Step-by-step explanation:

So the cashier counts 80 bills and if each bill was 10 dollars, then that would equal the total. ($800) Because 80 x 10 equals 800.

Hope this helps!. :)

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−15x+60≤105 AND14x+11≤−31
Mice21 [21]

Answer:

-15x<=45 and 14x<=-42

x>=45/-15 and x<=-42/14

x>=-3 and x<=-3

so x=-3

Step-by-step explanation:

6 0
3 years ago
V
Andru [333]

Answer: 6 to 48

Step-by-step explanation:

1 x 6 = 6

8 x 6 = 48

(pls mark me brainliest)

4 0
3 years ago
Which of the following are steps in practical problem solving?
finlep [7]

Answer:

Assign an identifying variable to the quantity to be found. Write a sentence stating conditions placed on the quantity. Solve the sentence for the variable.

Step-by-step explanation:

can you mark me as brainlist

8 0
2 years ago
Find the roots of the equation f(x) = x3 - 0.2589x2 + 0.02262x -0.001122 = 0
devlian [24]

Answer:

The root of the equation x^3-0.2589x^{2}+0.02262x-0.001122=0 is x ≈ 0.162035

Step-by-step explanation:

To find the roots of the equation x^3-0.2589x^{2}+0.02262x-0.001122=0 you can use the Newton-Raphson method.

It is a way to find a good approximation for the root of a real-valued function f(x) = 0. The method starts with a function f(x) defined over the real numbers, the function derivative f', and an initial guess x_{0} for a root of the function. It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it.

This is the expression that we need to use

x_{n+1}=x_{n} -\frac{f(x_{n})}{f(x_{n})'}

For the information given:

f(x) = x^3-0.2589x^{2}+0.02262x-0.001122=0\\f(x)'=3x^2-0.5178x+0.02262

For the initial value x_{0} you can choose x_{0}=0 although you can choose any value that you want.

So for approximation x_{1}

x_{1}=x_{0}-\frac{f(x_{0})}{f(x_{0})'} \\x_{1}=0-\frac{0^3-0.2589\cdot0^2+0.02262\cdot 0-0.001122}{3\cdot 0^2-0.5178\cdot 0+0.02262} \\x_{1}=0.0496021

Next, with x_{1}=0.0496021 you put it into the equation

f(0.0496021)=(0.0496021)^3-0.2589\cdot (0.0496021)^2+0.02262\cdot 0.0496021-0.001122 = -0.0005150, you can see that this value is close to 0 but we need to refine our solution.

For approximation x_{2}

x_{2}=x_{1}-\frac{f(x_{1})}{f(x_{1})'} \\x_{1}=0-\frac{0.0496021^3-0.2589\cdot 0.0496021^2+0.02262\cdot 0.0496021-0.001122}{3\cdot 0.0496021^2-0.5178\cdot 0.0496021+0.02262} \\x_{1}=0.168883

Again we put x_{2}=0.168883 into the equation

f(0.168883)=(0.168883)^3-0.2589\cdot (0.168883)^2+0.02262\cdot 0.168883-0.001122=0.0001307 this value is close to 0 but again we need to refine our solution.

We can summarize this process in the following table.

The approximation x_{5} gives you the root of the equation.

When you plot the equation you find that only have one real root and is approximate to the value found.

5 0
3 years ago
Help me do 7,8 and 9 please
kogti [31]
YOU HAVE TO FIND THE SLOPE m

The formula to find the slope is m=(y₂-y₁)/(x₂-x₁)

Since we have the coordinate of each point, then we can calculate the slope m

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7 0
4 years ago
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