17*7! Will mark brainliest

Which of the following is the point and slope of the equation y + 5 = 3(x + 9)?

Tatiana [17]

Answer:

Option B, (-9, -5), 3

Step-by-step explanation:

The point slope form in: y - y1 = m(x - x1)

m is the slope

(x1, y1) is the point

Step 1: Determine what the slope and point is their equation

y + 5 = 3(x + 9)

3 is the slope

(-9, -5) is the point

Answer: Option B, (-9, -5), 3

3 0

3 years ago

1360 concert tickets were sold for a total of $13,328. If students paid $8 and non students paid $12, how many student tickets w

lyudmila [28]

748 student tickets were sold.

Step-by-step explanation:

Given,

Number of tickets sold = 1360

Revenue generated = $13328

Cost of each student ticket = $8

Cost of each non student ticket = $12

Let,

x be the number of student tickets sold

y be the number of non student tickets sold

According to given statement;

x+y=1360 Eqn 1

8x+12y=13328 Eqn 2

Multiplying Eqn 1 by 12

$12(x+y=1360)\\12x+12y=16320\ \ \ Eqn\ 3$

Subtracting Eqn 2 from Eqn 3

$(12x+12y)-(8x+12y)=16320-13328\\12x+12y-8x-12y=2992\\4x=2992\\$

Dividing both sides by 4

$\frac{4x}{4}=\frac{2992}{4}\\x=748$

748 student tickets were sold.

Keywords: linear equation, elimination method

Learn more about elimination method at:

#LearnwithBrainly

5 0

3 years ago

15 _2_3 = 9 Fill in the gaps from left to right.

ruslelena [56]

15-2x3=9 following the rule of PEMDAS

8 0

3 years ago

Read 2 more answers

PLEASE HELP MEEEEEEEEEEEEEEEEE ASAAAAAAAP How many centimeters are in 10 meters? A) 10,000 cm B) 1,000 cm C) 100 cm D) 10 cm

statuscvo [17]

Answer:

1,000 cm......there are 100 cm per meter so 1,000 cm per 10 meters

Step-by-step explanation:

8 0

4 years ago

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

3 years ago