Find p and q, if Definition area is [3;10]

Mathematics

1 answer:

miss Akunina [59]2 years ago

5 0

$\\ \tt\longmapsto y=\sqrt{-x^2+px+q}$

Put(3,10)

$\\ \tt\longmapsto y^2=-x^2+px+q$

$\\ \tt\longmapsto 10^2=-(3^)2+3p+q$

$\\ \tt\longmapsto 100=-9+3p+q$

$\\ \tt\longmapsto 3p+q=109$

Use hit and trial method.

Let p,q be (27,28)

Apply

$\\ \tt\longmapsto 3(27)+28=81+28=109$

Verified

You might be interested in

A helicopter flies 665 feet above ground. What is the measure of the angle of depression from the helicopter to the girl’s line

grigory [225]

67 is the correct answer trust me i just answered and got it correct.

8 0

3 years ago

Read 2 more answers

Find the value of 35° 9

Liono4ka [1.6K]

Answer:

26°9

iuuiiii not sure tttt

8 0

2 years ago

Let f(x)=(18)3^x. find the equation of the line between the points (-2,f(-2)) and (1,f(1))

horsena [70]

$\text{Given that,} \\\\f(x) = 18 \cdot 3^x \\\\f(-2) = 18 \cdot 3^{-2} = 2\\\\f(1) = 18 \cdot 3^1 = 54\\ \\\text{So,}~ (x_1,y_1) = (x_1, f(-2)) = (-2,2) ~\text{and}~ (x_2,y_2) = (x_2, f(1)) = (1,54)\\ \\\text{Slope,}~ m = \dfrac{y_2 -y_1}{x_2 -x_1} = \dfrac{54-2}{1+2} = \dfrac{52}3 \\\\$

$\text{Equation of line,}\\\\~~~~~~~y - y_1 = m (x-x_1)\\\\\\\implies y-2 = \dfrac{52}3(x+2)~~~~~~~~~~~;[\text{Point-Slope form.]} \\\\\\\implies y-2 = \dfrac{52}3x + \dfrac{104}3\\\\\\\implies y = \dfrac{52}{3}x + \dfrac{104} 3 +2\\\\\\\implies y = \dfrac{52}3 x +\dfrac{110}3 ~~~~~~~~~~~~~~~;[\text{Slope -Intercept form}]$