All categories

2 years ago

If PQRS is a rhombus, which statements must be true? Check all that apply.

Mathematics

1 answer:

Alexxx [7]2 years ago

6 0

Answer:

PT=RT

PR is prependicular to qs

PQ=QR

QPR is congruent to SPR

Step-by-step explanation:

You might be interested in

Hello please help posted picture of question

Lynna [10]

Answer:

A. True

Step-by-step explanation:

All function of the form y = a^x, where a is a positive constant, goes through the point (0,1) and y = 0 is the asymptote.

Also note, a function y = a^x + k is a vertically translated function from the parent (y=a^x) k units up.

So, the function F(x) = 7 + 3^x is basically 3^x translated 7 units. What happens is that the point (0,1) gets translated 7 units above, so this point becomes (0,1+7)=(0,8). and the asymptote is y = 7.

Hence the function starts from above y = 7 (y = 7 is asymptote), goes through (0,8) and increases. Hence the Range would be all numbers GREATER than 7 (till infinity).

So, the question is TRUE.

4 0

2 years ago

6+9-11:44-223=x<br> lolshhsskzhzhzhxu

nadezda [96]

Answer:

what up my guy?

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

HURRY THIS IS REALLY IMPORTANT

charle [14.2K]

Answer:\ b.

Step-by-step explanation:

Hope this helps!!

5 0

2 years ago

Read 2 more answers

A restaurant operator in Accra has found out that during the lockdown,if she sells a plate of her food for Ghc 20 each,she can s

ella [17]

The demand equation illustrates the price of an item and how it relates to the demand of the item.

The slope of the demand function is -1/2
The equation of the demand function is: $R(x) = (300 - 10x) \times (20 + 5x)$
The price that maximizes her revenue is: Ghc 85

From the question, we have:

$Plates = 300$

$Price = 20$

The number of plates (x) decreases by 10, while the price (y) increases by 5. The table of value is:

$\begin{array}{cccccc}x & {300} & {290} & {280} & {270} & {260} \ \\ y & {20} & {25} & {30} & {35} & {40} \ \end{array}$

The slope (m) is calculated using:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

So, we have:

$m = \frac{25-20}{290-300}$

$m = \frac{5}{-10}$

$m = -\frac{1}{2}$

The equation of the demand is as follows:

The initial number of plates (300) decreases by 10 is represented as: (300 - 10x).

Similarly, the initial price (20) increases by 5 is represented as: (20 + 5x).

So, the demand equation is:

$R(x) = (300 - 10x) \times (20 + 5x)$

Open the brackets to calculate the maximum revenue

$R(x) =6000 + 1500x - 200x - 50x^2$

$R(x) =6000 + 1300x - 50x^2$

Equate to 0

$6000 + 1300x - 50x^2 =0$

Differentiate with respect to x

$1300 - 100x =0$

Collect like terms

$100x =1300$

Divide by 100

$x =13$

So, the price at maximum revenue is:

$Price= 20 + 5x$

$Price= 20 + 5 * 13$

$Price= 85$

In conclusion:

The slope of the demand function is -1/2
The equation of the demand function is: $R(x) = (300 - 10x) \times (20 + 5x)$
The price that maximizes her revenue is: Ghc 85