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3 years ago

Please help with functions

Mathematics

1 answer:

OlgaM077 [116]3 years ago

8 0

Answer:

f(0)=0

Step-by-step explanation:

The number in the parenthesis in a function are what you substitute for x, so do that for the problem:

f(x)=(1/2)(x) --> f(0)=(1/2)(0)

Then solve for f(0) by simplifying the other side of the equation:

f(0)=0

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The school that Jaidee goes to is selling tickets to the annual talent show. On the first day of ticket sales the school sold 10

Olenka [21]

The price of a senior citizen ticket is $8 and the price of a student ticket is$ 10

Step-by-step explanation:

Let the cost of one seniors citizen ticket be x and student be y

First day:

10 seniors citizen tickets and 11 student for a total of $190.

10x + 11y = 190-----------------------------(1)

Second day:

$160 on the second day by selling 5 senior citizen tickets and 12 student tickets.

5x + 12y = 160---------------------------------(2)

Multiply eq(2) by 2

10x + 24y = 320----------------------------(3)

Subtract eq(1) from (3)

10x + 24y = 320

10x + 11y = 190

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

0x + 13 y = 130

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

13y = 130

$y = \frac{130}{13}$

y = 10-------------------------------(4)

Substituting (4) in (2)

5x + 12(10) = 160

5x + 120 = 160

5x = 160-120

5x = 40

$x = \frac{40}{5}$

x = 8

5 0

3 years ago

ASAP 30 + Brainliest Please only solve 2 - 5

hichkok12 [17]

QUESTION 2a

We want to find the area of the given right angle triangle.

We use the formula

$Area=\frac{1}{2}\times base\times height$

The height of the triangle is $=a cm$ .

The base is $12cm$ .

We substitute the given values to obtain,

$Area=\frac{1}{2}\times 12\times a cm^2$ .

This simplifies to get an expression for the area to be

$Area=6a cm^2$ .

QUESTION 2b

The given diagram is a rectangle.

The area of a rectangle is given by the formula

$Area=length \times width$

The length of the rectangle is $l=7cm$ and the width of the rectangle is $w=ycm$ .

We substitute the values to obtain the area to be

$Area=7 \times y$

The expression for the area is

$Area=7y$

QUESTION 2c.

The given diagram is a rectangle.

The area of a rectangle is given by the formula

$Area=length \times width$

The length of the rectangle is $l=2x cm$ and the width of the rectangle is $w=4 cm$ .

We substitute the values to obtain the area to be

$Area=2x \times 4$

The expression for the area is

$Area=8x$

QUESTION 2d

The given diagram is a square.

The area of a square is given by,

$Area=l^2$ .

where $l=b m$ is the length of one side.

The expression for the area is

$Area=b^2 m^2$

QUESTION 2e

The given diagram is an isosceles triangle.

The area of this triangle can be found using the formula,

$Area=\frac{1}{2}\times base\times height$ .

The height of the triangle is $4cm$ .

The base of the triangle is $6a cm$ .

The expression for the area is

$Area=\frac{1}{2}\times 6a \times 4cm^2$

$Area=12a cm^2$

QUESTION 3a

Perimeter is the distance around the figure.

Let P be the perimeter, then

$P=x+x+x+x$

The expression for the perimeter is

$P=4x mm$

QUESTION 3b

The given figure is a rectangle.

Let P, be the perimeter of the given figure.

$P=L+B+L+B$

This simplifies to

$P=2L+2B$

$P=2(L+B)$

QUESTION 3c

The given figure is a parallelogram.

Perimeter is the distance around the parallelogram

$Perimeter=3q+P+3q+P$

This simplifies to,

$Perimeter=6q+2P$

$Perimeter=2(3q+P)$

QUESTION 3d

The given figure is a rhombus.

The perimeter is the distance around the whole figure.

Let P be the perimeter. Then

$P=5b+5b+5b+5b$

This simplifies to,

$P=20b mm$

QUESTION 3e

The given figure is an equilateral triangle.

The perimeter is the distance around this triangle.

Let P be the perimeter, then,

$P=2x+2x+2x$

We simplify to get,

$P=6x mm$

QUESTION 3f

The figure is an isosceles triangle so two sides are equal.

We add all the distance around the triangle to find the perimeter.

This implies that,

$Perimeter=3m+5m+5m$

$Perimeter=13m mm$

QUESTION 3g

The given figure is a scalene triangle.

The perimeter is the distance around the given triangle.

Let P be the perimeter. Then

$P=(3x+1)+(2x-1)+(4x+5)$

This simplifies to give us,

$P=3x+2x+4x+5-1+1$

$P=9x+5$

QUESTION 3h

The given figure is a trapezium.

The perimeter is the distance around the whole trapezium.

Let P be the perimeter.

Then,

$P=m+(n-1)+(2m-3)+(n+3)$

We group like terms to get,

$P=m+2m+n+n-3+3-1$

We simplify to get,

$P=3m+2n-1$ mm

QUESTION 3i

The figure is an isosceles triangle.

We add all the distance around the figure to obtain the perimeter.

Let $P$ be the perimeter.

Then $P=(2a-b)+(a+2b)+(a+2b)$

We regroup the terms to get,

$P=2a+a+a-b+2b+2b$

This will simplify to give us the expression for the perimeter to be

$P=4a+3b$ mm.

QUESTION 4a

The given figure is a square.

The area of a square is given by the formula;

$Area=l^2$

where $l=2m$ is the length of one side of the square.

We substitute this value to obtain;

$Area=(2m)^2$

This simplifies to give the expression of the area to be,

$Area=4m^2$

QUESTION 4b

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=5a cm$ is the length of the rectangle and $w=6cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =5a \times 6$

$Area =30a cm^2$

QUESTION 4c

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=7y cm$ is the length of the rectangle and $w=2x cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =7y \times 2x$

The expression for the area is

$Area =14xy cm^2$

QUESTION 4d

The given figure is a rectangle.

The formula for finding the area of a rectangle is

$Area=l\times w$ .

where $l=3p cm$ is the length of the rectangle and $w=p cm$ is the width of the rectangle.

We substitute the values into the formula to get,

$Area =3p \times p$

The expression for the area is

$Area =3p^2 cm^2$

See attachment for the continuation

6 0

3 years ago

Read 2 more answers

When e = 4, f = 2, and g = 8If e varies jointly with f and g, what is the constant of variation?

tatyana61 [14]

Answer:

$e=\frac{1}{4} fg$

Step-by-step explanation:

Joint variation problem solve using the equation y = kxz.

e ∝ fg

e=kfg

now substitute the values

$4=k*2*8\\4=16k\\k=\frac{4}{16} \\k=\frac{1}{4}$

The relationship will be:

$e=kfg\\e=\frac{1}{4} fg$

8 0

3 years ago

Read 2 more answers

Line DA is tangent to circle B. What is the measure of angle ACB?

alexdok [17]

A tangent line creates a 90 degree angle with a radius line that touches the tangent Line.

Line BC is a radius that touches the tangent line at point C

The angle created is 90 degrees.

The answer is C. 90 degrees.

6 0

3 years ago

Read 2 more answers

The weights (to the nearest pound) of some boxes to be shipped are found to be:. Weight 65 68 69 70 71 72 90 95 frequency 1 2 5

Lena [83]

So the mean is 72.97

We need to subtract the mean from each value and square it.
(65-72.97)^2= 63.5209
(68-72.97)^2=24.7009
(69-72.97)^2=15.7609
(70-72.97)^2=8.8209
(71-72.97)^2= 3.8809
(72-72.97)^2=0.9409
(90-72.97)^2=290.0209
(95-72.97)^2=485.3209

Now we add up the new values ( also consider their frequency) and find their mean.
Add the values
63.5209+(2 •24.7009=49.4018)+(5•15.7609=78.8045)+(8•8.8209=70.5672)+(7•3.8809=27.1663)+(3•0.9409=2.8227)+(2•290.0209=580.0418)+(2•485.3209=970.6418)= 1,842.967
Divide by total numburs to find the mean
1,842.967/ 30=61.43223333

The standar deviation is the square root of the mean so is
Square root of 61.43223333=7.837871735
Round to the nearest tenth
Standard Deviation is 7.8

7 0

3 years ago