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2 years ago

7

The high-school football team scored 19 points on touchdowns (worth 7 points if you count the extra point) and field goals (wort

h 3 points). If the team scored 5 times, how many of each type did they score?

1 answer:

Kazeer [188]2 years ago

5 0

If they scored 5 times, they got 1 touchdown and 4 field goals.

Equation:

7 + 3 + 3 + 3 + 3 = 19.

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2 years ago

The sum of four times a number and five times another number is -28. The sum of nine times the first number and two times the se

Rina8888 [55]

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<h2>3 and -8</h2>

Step-by-step explanation:

$x,\ y-the\ numbers\\\\\text{The sum of four times a number and five times another number is -28:}\\\\4x+5y=-28\\\\\text{The sum of nine times the first number and two times the second number is 11:}\\\\9x+2y=11\\\\\text{We have the system of equations:}\\\\\left\{\begin{array}{ccc}4x+5y=-28&\text{multiply both sides by2}\\9x+2y=11&\text{multiply both sides by (-5)}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}8x+10y=-56\\-45x-10y=-55\end{array}\right}\qquad\text{add both sides of the equation}$

$.\qquad-37x=-111\qquad\text{divide both sides by (-37)}\\.\qquad\qquad\boxed{x=3}\\\\\text{Put the value of x to the first equation:}\\\\4(3)+5y=-28\\12+5y=-28\qquad\text{subtrac 12 from both sides}\\5y=-40\qquad\text{divide both sides by 5}\\\boxed{y=-8}$

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2 years ago

What is the positive slope of the asymptote of (y+11)^2/100-(x-6)^2/4=1? The positive slope of the asymptote is .

VladimirAG [237]

Answer:

the positive slope of the asymptote = 5

Step-by-step explanation:

Given that:

$\frac{(y+11)^2}{100} -\frac{(x-6)^2}{4} =1$

Using the standard form of the equation:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}= 1$

where:

(h,k) are the center of the hyperbola.

and the y term is in front of the x term indicating that the hyperbola opens up and down.

a = distance that indicates how far above and below of the center the vertices of the hyperbola are.

For the above standard equation; the equation for the asymptote is:

$y = \pm \frac{a}{b} (x-h)+k$

where;

$\frac{a}{b}$ is the slope

From above;

(h,k) = 11, 100

$a^2$ = 100

a = $\sqrt{100}$

a = 10

$b^2 =4$

b = $\sqrt{4}$

b = 2

$y = \pm \frac{10}{2} (x-11)+2$

$y = \pm 5 (x-11)+2$

y = 5x-53 , -5x -57

Since we are to find the positive slope of the asymptote: we have

$\frac{a}{b}$ to be the slope in the equation $y = \pm \frac{10}{2} (y-11)+2$

$\frac{a}{b}$ = $\frac{10}{2}$

$\frac{a}{b}$ = 5

Thus, the positive slope of the asymptote = 5

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2 years ago