MATHEMATICS HIGH SCHOOL

The question is in the picture.

Answers

Answer 1

Answer:

Answer:

The third answer choice

Step-by-step explanation:

So sqrt(5)-sqrt(9*5) or sqrt(5)-3sqrt(5). We can rewrite this as -2sqrt5.

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Find the sum of - 3x^2 + 5x – 8 and -10x^2 – X – 3.

Answers

The sum of - $3x^2 + 5x - 8$ and $-10x^2 - x - 3$ is: $-13x^2 + 4x - 11.$

To find the sum of $-3x^2 + 5x - 8$ and $-10x^2 - x - 3,$ you simply add the like terms together.

Like terms are terms that have the same variable and the same exponent.

$-3x^2 + 5x - 8$

$(-10x^2 - x - 3)$

Now, add the like terms:

$-3x^2 + (-10x^2) = -13x^2$

5x + (-x) = 4x

-8 + (-3) = -11

For similar question on sum.

brainly.com/question/24205483

#SPJ3

Answer:

-13x^2+4x-11

Step-by-step explanation:

-3x^2+5x-8+(-10x^2-x-3)=?

-3x^2+5x-8-10x^2-x-3=?

(combine like terms); -13x^2+4x-11

So, your answer is -13x^2+4x-11

The vertex of this parabola is at (-3, 6). Which of the following could be its equation? A. y = -3(x + 3)2 + 6
B. y = -3(x + 3)2 - 6
C. y = -3(x - 3)2 + 6
D. y = -3(x - 3)2 - 6

Answers

nice, already in vertex form
y=a(x-h)^2+k
(h,k) is vertex

therfor since (-3,6) is vertex
we are looking for something like
y=a(x-(-3))^2+6 simplified to
y=a(x+3)^2+6

A is ansre

So,

The vertex form of a parabola is:

$y=a(x-h)^2+k$

and the vertex is at (h,k).

Thus, the equation will have to look something like this:

$y=a(x+3)^2+6$

Only option A matches these criteria, so is must be the correct option.

The function y = -0.03(x - 14)^2 + 6 models the mump of a red kangaroo where x is the horizontal distance in meters and y is the vertical distance in meters for the height of the jump. What is the kangaroo's maximum height? How long is the kangaroo's jump?

Answers

From the vertex of the quadratic equation, we find that: The kangaroos maximum height is of 6 meters.

From the roots of the equation, we find that: The kangaroo's jump is 28.14 meters long.

----------------------------

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

$f(x) = ax^(2) + bx + c$

It's vertex is the point $(x_(v), y_(v))$

In which

$x_(v) = -(b)/(2a)$

$y_(v) = -(\Delta)/(4a)$

Where

$\Delta = b^2-4ac$

If a<0, the vertex is a maximum point, that is, the maximum value happens at $x_(v)$ , and it's value is $y_(v)$ .

----------------------------

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^(2) + bx + c, a\neq0$ .

This polynomial has roots $x_(1), x_(2)$ such that $ax^(2) + bx + c = a(x - x_(1))*(x - x_(2))$ , given by the following formulas:

$x_(1) = (-b + √(\Delta))/(2*a)$

$x_(2) = (-b - √(\Delta))/(2*a)$

$\Delta = b^(2) - 4ac$

----------------------------

The quadratic equation is:

$y = -0.03(x - 14)^2 + 6$

Placing in standard form:

$y = -0.03(x^2 - 28x + 196) + 6$

$y = -0.03x^2 + 0.84x + 0.12$

Thus, it has coefficients $a = -0.03, b = 0.84, c = 0.12$

----------------------------

The kangaroo's maximum height is the y-value of the vertex, thus:

$\Delta = b^2 - 4ac = (0.84)^2 - 4(-0.03)(0.12) = 0.72$

$y_(v) = -(\Delta)/(4a) = -(0.72)/(4(-0.03)) = 6$

The kangaroos maximum height is of 6 meters.

----------------------------

The length of the kangaroo's jump is the positive root. The roots are found at the values of x for which y = 0, thus, the solutions of the quadratic equation.

$x_(1) = (-0.84 + √(0.72))/(2(-0.03)) = -0.14$

$x_(2) = (-0.84 - √(0.72))/(2(-0.03)) = 28.14$

The kangaroo's jump is 28.14 meters long.

A similar question is given at brainly.com/question/16858635

Answer:

Kangaroo's maximum height is 6 m and the kangaroo's jump is 28 m long

Step-by-step explanation:

Given : $y = -0.03(x - 14)^2 + 6$

To Find : What is the kangaroo's maximum height? How long is the kangaroo's jump?

Solution:

$y = -0.03(x - 14)^2 + 6$

x is the horizontal distance in meters

y is the vertical distance in meters for the height of the jump.

Substitute y = 0

$0 = -0.03(x - 14)^2 + 6$

$0.03(x - 14)^2 = 6$

$(x - 14)^2 = (6)/(0.03)$

$(x - 14)^2 =200$

$(x - 14) =√(200)$

$(x - 14) =14.142$

$x =14.142+14$

$x =28.142$

x≈ 28 m

Now the maximum height will be attained at mid point i.e. $(28)/(2) =14$

Now substitute x= 14

$y = -0.03(14 - 14)^2 + 6$

$y = 6$

So, kangaroo's maximum height is 6 m and the kangaroo's jump is 28 m long

Complete the square
t^2+10t=75

Answers

COMPLETING THE SQUARE (PROCESS):
Do a reverse factoring procedure to develop the following form of equation:
(t + a)^2 = b

where:
t = unknown variable
a = coefficient
b = coefficient

Then take square root of both sides to find value of unknown variable as follows:

t^2 + 10t = 75
(t^2 + 10t + 25) = 75 + 25
(t^2 + 5t + 5t + 25) = 100
(t + 5)(t + 5) = 100
(t + 5)^2 = 100
√(t + 5)^2 = √100

Two solutions exist:
+(t + 5) = 10
AND
-(t + 5) = 10

Thus:
t = 10 - 5 = 5
AND
t = -5 - 10 = -15

Answer:
t = 5
AND
t = -15

$t^2+10t=75 \n t^2+10t+25-25=75\n (t+5)^2=100\n |t+5|=10\n t+5=10 \vee t+5=-10\n t=5 \vee t=-15$

the suntracker grows at a rate of 2.5 CM per day after the first 60 days.If this sunflower is 195 cm tall when it is 60 days old witer a expression to represent suntrackers height after 22 days or when it is 82 days old exspain how you found your answer, other sunflower is 235 cm

Answers

Given:
height at 60 days old - 195cm
height after 60 days old - 2.5cm per day

total height = 195 cm + 2.5cm(x - 60)
x = day measured

22 days old

total height = 195cm + 2.5cm(22-60)
t.h = 195cm + 2.5cm(-38)
t.h = 195cm - 95cm
t.h = 100 cm

82 days old

total height = 195cm + 2.5cm(x-60)
t.h = 195cm + 2.5cm(82-60)
t.h = 195cm + 2.5cm(22)
t.h = 195cm + 55cm
t.h = 250cm

You are going on a 1,940-mile trip, and your car gets about 28 miles per gallon. Gas prices along your route average $2.95 per gallon. Calculate the cost of thetrip. Round final answer to the nearest cent.

Answers

Answer: $204.39

Reason

The number of gallons used is estimated to be approximately: distance/mpg = 1940/28 = 69.2857 gallons.

Each gallon costs $2.95, so the total cost is 69.2857*2.95 = 204.392815 which rounds to 204.39

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The point-slope form of the equation of the line that passes through (–4, –3) and (12, 1) is y – 1 = (x – 12). What is the standard form of the equation for this line?x – 4y = 8 x – 4y = 2 4x – y = 8 4x – y = 2