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4 years ago

-x+y=4 3x-2y=-7 substitution

1 answer:

4 0

-x+y=4 add x to both sides to get y=x+4 Plug x+4 into the equation for y to get 3x-2(x+4)=-7 Distribute to get 3x-2x-8=-7 Add 8 to both sides to get 3x-2x=1 Subtract 3x-2x to get x=1 the answer is 1

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Could you help me to solve the problem below the cost for producing x items is 50x+300 and the revenue for selling x items is 90

s344n2d4d5 [400]

Answer:

Profit function: P(x) = -0.5x^2 + 40x - 300

profit of $50: x = 10 and x = 70

NOT possible to make a profit of $2,500, because maximum profit is $500

Step-by-step explanation:

(Assuming the correct revenue function is 90x−0.5x^2)

The cost function is given by:

$C(x) = 50x + 300$

And the revenue function is given by:

$R(x) = 90x - 0.5x^2$

The profit function is given by the revenue minus the cost, so we have:

$P(x) = R(x) - C(x)$

$P(x) = 90x - 0.5x^2 - 50x - 300$

$P(x) = -0.5x^2 + 40x - 300$

To find the points where the profit is $50, we use P(x) = 50 and then find the values of x:

$50 = -0.5x^2 + 40x - 300$

$-0.5x^2 + 40x - 350 = 0$

$x^2 - 80x + 700 = 0$

Using Bhaskara's formula, we have:

$\Delta = b^2 - 4ac = (-80)^2 - 4*700 = 3600$

$x_1 = (-b + \sqrt{\Delta})/2a = (80 + 60)/2 = 70$

$x_2 = (-b - \sqrt{\Delta})/2a = (80 - 60)/2 = 10$

So the values of x that give a profit of $50 are x = 10 and x = 70

To find if it's possible to make a profit of $2,500, we need to find the maximum profit, that is, the maximum of the function P(x).

The maximum value of P(x) is in the vertex. The x-coordinate of the vertex is given by:

$x_v = -b/2a = 80/2 = 40$

Using this value of x, we can find the maximum profit:

$P(40) = -0.5(40)^2 + 40*40 - 300 = $500$

The maximum profit is $500, so it is NOT possible to make a profit of $2,500.

3 0

3 years ago

In expanded form, in kilometers, how far is Jupiter from the Sun?

DIA [1.3K]

<span>740,679,835 km i hope this helps you
</span>

5 0

3 years ago

Dairy farmers are aware there is often a linear relationship between the age, in years, of a dairy cow and the amount of milk pr

Jobisdone [24]

Answer:

B. A cow of 5 years is predicted to produce 5.5 more gallons per week.

Step-by-step explanation:

Let $M(a) = 40.8-1.1\cdot a$ , where $a$ is the age of the dairy cow, measured in years, and $M(a)$ is the predicted milk production, measured in gallons per week.

Besides, we consider $a_{1}$ and $a_{2}$ , such that $a_{1}\ne a_{2}$ , we define the difference between predicted milk productions ( $\Delta M$ ) below:

$\Delta M = -1.1\cdot (a_{2}-a_{1})$ (1)

If we know that $a_{1} = 5\,yr$ and $a_{2} = 10\,yr$ , then the difference between predicted milk productions is:

$\Delta M = -1.1\cdot (10-5)$

$\Delta M = -5.5\,\frac{gal}{week}$

That is, a cow of 5 years is predicted to produce 5.5 more gallons per week than a cow of 10 years. Hence, the right answer is B.

6 0

3 years ago

The question is on the image

allochka39001 [22]

2th bounce- 80
5th Bounce- (-)200
9th Bounce- (-)560

Every time the ball bounces it loses 40cm because on the first bounce 160cm-120cm=40cm
And the score goes down by 40 every time and since we go into the negatives the negatives sometimes add on top of each other

5 0

2 years ago

Which of the following is not a property of a chi-square distribution?

laiz [17]

Answer:

c) Is not a property (hence (d) is not either)

Step-by-step explanation:

Remember that the chi square distribution with k degrees of freedom has this formula

$\chi_k^2 = \matchal{N}_1^2 + \matchal{N}_2^2 + ... + \, \matchal{N}_{k-1}^2 + \matchal{N}_k^2$

Where N₁ , N₂m .... $N_k$ are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.

Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true

The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.

6 0

3 years ago