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3 years ago

What is the area of the parallelogram?

Mathematics

1 answer:

nevsk [136]3 years ago

7 0

The length of the parallelogram is 3/4 ft less 1/8 ft, or 6/8 ft less 1/8 ft, or 5/8 ft.

The width is 1/2 ft.

Thus, the area is (5/8 ft)(1/2 ft) = (5/16) ft^2

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Step-by-step explanation:

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3 years ago

Find a function of the form y=ax^2+bx+c whose graph passes through (-11,2),(-9,-2)and (-5,14). (Vertex form also?) please help

nikitadnepr [17]

Answer:

$y=x^2+18x+79$

$y=(x+9)^2-2$

Step-by-step explanation:

We want to find a function of the form:

$y=ax^2+bx+c$

We know that it passes through the three points: (-11, 2); (-9, -2); and (-5,14).

In other words, if we substitute -11 for x, we should get 2 for y. So:

$(2)=a(-11)^2+b(-11)+c$

Simplify:

$2=121a-11b+c$

We will do it to the two other points as well. So, for (-9, -2):

$\begin{aligned}(-2)&=a(-9)^2+b(-9)+c\\\Rightarrow-2&=81a-9b+c\end{aligned}$

And similarly:

$\begin{aligned}(14)&=a(-5)^2+b(-5)+c\\\Rightarrow 14&=25a-5b+c\end{aligned}$

So, to find our function, we will need to determine the values of a, b, and c.

We essentially have a triple system of equations:

$\begin{cases}2=121a-11b+c\\-2=81a-9b+c\\14=25a-5b+c\end{cases}$

To approach this, we can first whittle it down only using two variables.

So, let's use the First and Second Equation. Let's remove the variable b. To do so, we can use elimination. We can multiply the First Equation by 9 and the Second Equation by -11. This will yield:

$\begin{cases}{\begin{aligned}9(2)&=9(121a-11b+c)\\-11(-2)&=-11(81a-9b+c)\end{cases}}$

Distribute:

$\begin{cases}18=1089a-99b+9c\\22=-891a+99b-11c\end{cases}$

Now, we can add the two equations together:

$(18+22)=(1089a-891a)+(-99b+99b)+(9c-11c)$

Simplify:

$40=198a-2c$

Now, let's do the same using the First and Third Equations. We want to cancel the variable b. So, let's multiply the First Equation by 5 and the Third Equation by -11. So:

$\begin{cases}{\begin{aligned}5(2)=5(121a-11b+c)\\-11(14)=-11(25a-5b+c)\end{cases}$

Simplify:

$\begin{cases}10=605a-55b+5c\\-154=-275a+55b-11c\end{cases}$

Now, let's add the two equations together:

$(10-154)=(605a-275a)+(-55b+55b)+(5c-11c)$

Simplify:

$-144=330a-6c$

Therefore, we now have the two equations:

$\begin{cases}40=198a-2c\\-144=330a-6c\end$

Let's cancel the c. So, multiply the First Equation by -3. We don't have to do anything special to the second. So:

$-3(40)=-3(198a-2c)$

Multiply:

$-120=-594a+6c$

Now, add it to the Second Equation:

$(-120+-144)=(-594a+330a)+(6c-6c)$

Add:

$-264=-264a$

Divide both sides by -264. So, the value of a is:

$a=1$

Now, we can use either of the two equations above to obtain c. Let's use the first one. So:

$40=198a-2c$

Substitute 1 for a:

$40=198(1)-2c$

Solve for c. Subtract 198 from both sides and divide by -2:

$\begin{aligned}40&=198-2c\\ -158&=-2c \\79 &=c \end{aligned}$

So, the value of c is 79.

Finally, we can find b. We can use any of the three original equations. Let's use the First Equation. So:

$2=121a-11b+c$

Substitute 1 for a and 79 for c and determine the value of b:

$\begin{aligned}2&=121(1)-11b+(79)\\2&=121+79-11b\\2&=200-11b\\-198&=-11b\\18&=b\end{cases}$

Therefore, the value of b is 18.

So, our equation is:

$y=ax^2+bx+c$

Substitute in the values:

$y=(1)x^2+(18)x+(79)$

Simplify:

$y=x^2+18x+79$

Now, let's put this into vertex form. To do so, we will need to complete the square. First, let's group the first two terms together:

$y=(x^2+18x)+79$

To complete the square, we will divide the b term by 2 and then square it.

b is 18. 18/2 is 9 and 9² is 81. Therefore, we will add 81 into our parentheses:

$y=(x^2+18x+81)+79$

Since we added 81, we must also subtract 81. So:

$y=(x^2+18x+81)+79-81$

Subtract:

$y=(x^2+18x+81)-2$

The grouped terms are a perfect square trinomial. Factor:

$y=(x+9)^2-2$

And this is in vertex form.

And we are done!

6 0

3 years ago

ZDAC = ZBAD. What is the length of AB? Round to one decimal place.

goblinko [34]

Answer: Line AB = 5.2

Step-by-step explanation: We start with triangle ACD, with two sides given and angle A which shall be the reference angle can be calculated as,

SinA = opposite/hypotenuse

Where the opposite is 4.3 (line facing the reference angle) and the hypotenuse is 5.6 (line facing the right angle)

SinA = 4.3/5.6

SinA = 0.7679

By use of a calculator or a table of values

A = 50.17 degrees.

Having been told that both angles DAC and BAD are equal, then we move to triangle ADB where the reference angle is 50.17 (BAD) and the opposite is 4 (line facing the reference angle) and the unknown side is the hypotenuse (line AB).

Sin 50.17 = opposite/hypotenuse

Sin 50.17 = 4/AB

0.7679 = 4/AB

By cross multiplication we now have

AB = 4/0.7679

AB = 5.2090

Approximately to one decimal place,

AB = 5.2 units

6 0

4 years ago