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2 years ago

Write equations for each of the following graphs!NO LINKS!!

Mathematics

2 answers:

hammer [34]2 years ago

7 0

9514 1404 393

Answer:

y = √(x+4) -2
y = -(x -2)^2 -3

Step-by-step explanation:

The translation of a parent function f(x) by (h, k) gives the function ...

g(x) = f(x -h) +k . . . . . h units right, k units up

Reflection across the x-axis multiplies the function value by -1.

a) The square root function is translated 4 left and 2 down. The equation of the graph is ...

y = √(x +4) -2

b) The equation of the parabola is reflected across the x-axis and translated right 2 and down 3.

y = -(x -2)^2 -3

svetoff [14.1K]2 years ago

5 0

I would like to help you but I am doing the same and I do not understand him, but if you want we can help each other.

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A container is shaped like a triangle prism. Each base of container is an equilateral triangle with the dimensions shown. The co

Aleks04 [339]

Answer:

$\text{Lateral surface area of container}=270\text{ cm}^2$

Step-by-step explanation:

Please find the attachment.

We have been given that a container is shaped like a triangle prism. Each base of container is an equilateral triangle with each side 6 cm. The height of container is 15 cm.

To find the lateral surface area of our given container we will use lateral surface area formula of triangular prism.

$\text{Lateral surface area of triangular prism}=(a+b+c)*h$ , where, a, b and c represent base sides of prism and h represents height of the prism.

Upon substituting our given values in above formula we will get,

$\text{Lateral surface area of container}=(\text{6 cm+ 6 cm+6 cm})*\text{15 cm}$

$\text{Lateral surface area of container}=\text{18 cm}*\text{15 cm}$

$\text{Lateral surface area of container}=270\text{ cm}^2$

Therefore, lateral surface area of our given container is 270 square cm.

6 0

2 years ago

WILL MARK BRAINLIEST!!

Rudik [331]

Answer:

C = (A*P - 8.4Y -330T + 200I) / 100

Step-by-step explanation:

P = (8.4Y + 330T + 100C -200I ) / A

now we have to calculate completed passes C for given P, Y, T, I, A

A*P = 8.4Y + 330T -200I +100C

100C = A*P - 8.4Y - 330T + 200I

C = (A*P - 8.4Y -330T + 200I) / 100

I just solved the equation for C

5 0

2 years ago

A student is saving money to buy a microscope. The student currently has $50 and plans to save $15 every month. Write a function

Evgen [1.6K]

Answer:

Step-by-step explanation:

The student currently has $50 and plans to save $15 every month.

Let x represent the number of months that the student will save enough money to buy the microscope.

Let y represent the amount that the student saves after x months.

The function that represents the amount y (in dollars) of money that the student saves after x months will be

y = 50 + 15x

The 50 remains constant because she has already saved it

5 0

3 years ago

Komok [63]

Answer:

42.7 cm sq.

Step-by-step explanation:

small/large = 16 / 122

16/122 = S / 56

S = 56 * (16/122) = 7.34...

however not in the choice

ah ha..

You meant : 12.2 sq. cm : 16 sq. cm = small / large

so 12.2 / 16 = S / 56 cm

S = 42.7 sq. cm

8 0

3 years ago

Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given poi

lara [203]

For each curve, plug in the given point $(x,y)$ and check if the equality holds. For example:

(I) (2, 3) does lie on $x^2+xy-y^2=1$ since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.

For part (a), compute the derivative $\frac{\mathrm dy}{\mathrm dx}$ , and evaluate it for the given point $(x,y)$ . This is the slope of the tangent line at the point. For example:

(I) The derivative is

$x^2+xy-y^2=1\overset{\frac{\mathrm d}{\mathrm dx}}{\implies}2x+x\dfrac{\mathrm dy}{\mathrm dx}+y-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+y}{2y-x}$

so the slope of the tangent at (2, 3) is

$\dfrac{\mathrm dy}{\mathrm dx}(2,3)=\dfrac74$

and its equation is then

$y-3=\dfrac74(x-2)\implies y=\dfrac74x-\dfrac12$

For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, $-\frac1{\frac{\mathrm dy}{\mathrm dx}}$ . For example:

(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation

$y-3=-\dfrac47(x-2)\implies y=-\dfrac47x+\dfrac{29}7$

3 0

3 years ago

Write equations for each of the following graphs!NO LINKS!!​

Write equations for each of the following graphs!NO LINKS!!