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3 years ago

8

On a test, Andre gets 4 times as many answers correct as he gets wrong. Write the ratio of questions he gets correct to the tota

l number of questions.

2 answers:

STALIN [3.7K]3 years ago

5 0

Answer:

4:5

Step-by-step explanation:

NemiM [27]3 years ago

3 0

I think it's 1:4 since if he gets two wrong than two times four is eight .

You might be interested in

Solve 2x^2+x-4=0<br> X^2+_×+_=0

natima [27]

Answer:

Hopes it helps

Step-by-step explanation:

The Quadratic Polynomial is

2 x² +x -4=0

Using the Determinant method to find the roots of this equation

For, the Quadratic equation , ax²+ b x+c=0

(b) x²+x=0

x × (x+1)=0

x=0 ∧ x+1=0

x=0 ∧ x= -1

You can look the problem in other way

the two Quadratic polynomials are

2 x²+x-4=0, ∧ x²+x=0

x²= -x

So, 2 x²+x-4=0,

→ -2 x+x-4=0

→ -x -4=0

→x= -4

∨

x² +x² +x-4=0

x²+0-4=0→→x²+x=0

→x²=4

x=√4

x=2 ∧ x=-2

As, you will put these values into the equation, you will find that these values does not satisfy both the equations.

So, there is no solution.

You can solve these two equation graphically also.

3 0

3 years ago

Lin father is paying for a 20 meal he has a 15 off coupon for the meal what is the total after the discount

wel

Answer:

3

Step-by-step explanation:

20÷100=0.2

0.2×15=3

:)

4 0

2 years ago

Read 2 more answers

an electrician charges $50 to make a house call and $40 for each hour worked. if you have $200 can you afford a repair that take

mariarad [96]

No. you will be short 10$ so it will not work.

7 0

3 years ago

Read 2 more answers

Choose the most appropriate name for the function described below.

DedPeter [7]

T(p) would be an accurate answer. This situation likens to the f(x) function, where y is the function of x and so dependent on x. The cost of the tomatoes is dependent on the price per pound, so cost would be in the f place and price per pound would be in the (x) place.

4 0

3 years ago

Read 2 more answers

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 11z on the curve of intersection of the plane x − y + z =

Taya2010 [7]

Answer:

$\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}$

<em>Maximum value of f=2.41</em>

Step-by-step explanation:

<u>Lagrange Multipliers</u>

It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.

Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.

$\bigtriangledown f=\lambda \bigtriangledown g$

for some scalar $\lambda$ called the Lagrange multiplier.

For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is

$\bigtriangledown f=\lambda \bigtriangledown g+\mu \bigtriangledown h$

The gradient of f is

$\bigtriangledown f=$

Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in $x,y,z,\lambda,\mu$ .

We have

$f(x, y, z) = x + 2y + 11z\\g(x, y, z) = x - y + z -1=0\\h(x, y, z) = x^2 + y^2 -1= 0$

Let's compute the partial derivatives

$f_x=1\ ,f_y=2\ ,f_z=11\ \\g_x=1\ ,g_y=-1\ ,g_z=1\\h_x=2x\ ,h_y=2y\ ,h_z=0$

The Lagrange condition leads to

$1=\lambda (1)+\mu (2x)\\2=\lambda (-1)+\mu (2y)\\11=\lambda (1)+\mu (0)$

Operating and simplifying

$1=\lambda+2x\mu\\2=-\lambda +2y\mu \\\lambda=11$

Replacing the value of $\lambda$ in the two first equations, we get

$1=11+2x\mu\\2=-11 +2y\mu$

From the first equation

$\displaystyle 2\mu=\frac{-10}{x}$

Replacing into the second

$\displaystyle 13=y\frac{-10}{x}$

Or, equivalently

$13x=-10y$

Squaring

$169x^2=100y^2$

To solve, we use the restriction h

$x^2 + y^2 = 1$

Multiplying by 100

$100x^2 + 100y^2 = 100$

Replacing the above condition

$100x^2 + 169x^2 = 100$

Solving for x

$\displaystyle x=\pm \frac{10}{\sqrt{269}}$

We compute the values of y by solving

$13x=-10y$

$\displaystyle y=-\frac{13x}{10}$

For

$\displaystyle x= \frac{10}{\sqrt{269}}$

$\displaystyle y= -\frac{13}{\sqrt{269}}$

And for

$\displaystyle x= -\frac{10}{\sqrt{269}}$

$\displaystyle y= \frac{13}{\sqrt{269}}$

Finally, we get z using the other restriction

$x - y + z = 1$

Or:

$z = 1-x+y$

The first solution yields to

$\displaystyle z = 1-\frac{10}{\sqrt{269}}-\frac{13}{\sqrt{269}}$

$\displaystyle z = \frac{-23\sqrt{269}+269}{269}$

And the second solution gives us

$\displaystyle z = 1+\frac{10}{\sqrt{269}}+\frac{13}{\sqrt{269}}$

$\displaystyle z = \frac{23\sqrt{269}+269}{269}$

Complete first solution:

$\displaystyle x= \frac{10}{\sqrt{269}}\\\\\displaystyle y= -\frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{-23\sqrt{269}+269}{269}$

Replacing into f, we get

f(x,y,z)=-0.4

Complete second solution:

$\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}$

Replacing into f, we get

f(x,y,z)=2.4

The second solution maximizes f to 2.4

5 0

3 years ago