All categories

3 years ago

Please show work on how you got the answer

Mathematics

1 answer:

Maurinko [17]3 years ago

3 0

Since E is the midpoint, DE and EF are the same length.
Set the equation DE and EF equal to each other and solve for x.
2x + 4 = 3x - 1
x = 5
Then plug in 5 to find the length.
DE = 2(5) + 4 = 14
EF = 3(5) - 1 = 14
DF = DE + EF = 28

You might be interested in

A fruit seller purchased 20 kg of apples at Rs 80 per kg and sold them at Rs 85 per kg. find profit and profit percent

viktelen [127]

Answer:

Profit = Rs 100

Profit %age = 6.25%

Step-by-step explanation:

Cost Price = Rs 80

Cost Price for 20 kg = Rs 1600

Selling Price = Rs 85

Profit for 1 kg = Rs 85 - Rs 80

=> Rs 5 (For 1 Kg)

Profit for 20 kg:

=> Rs 5*20

=> Rs 100

Now Profit %age:

=> $\frac{100}{1600}* 100$

=> 0.0625 * 100

=> 6.25%

7 0

3 years ago

Read 2 more answers

How to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x

Fofino [41]

Answer:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

Step-by-step explanation:

We are given the function:

$f(x)=-4x^5-8x^3+12x$

And we want to finds its zeros.

Therefore:

$0=-4x^5-8x^3+12x$

Firstly, we can divide everything by -4:

$0=x^5+2x^3-3x$

Factor out an x:

$0=x(x^4+2x^2-3)$

This is in quadratic form. For simplicity, we can let:

$u=x^2$

Then by substitution:

$0=x(u^2+2u-3)$

Factor:

$0=x(u+3)(u-1)$

Substitute back:

$0=x(x^2+3)(x^2-1)$

By the Zero Product Property:

$x=0\text{ and } x^2+3=0\text{ and } x^2-1=0$

Solving for each case:

$x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}$

Therefore, our real and complex zeros are:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

4 0

2 years ago

Which equation represents this number sentence? Two more than one-third of a number is 8

Alja [10]

Answer:

Step-by-step explanation:

One - third = 1/3

then:

one third of a number = a*1/3 = a/3

Then the equation that represents this sentence is:

2 + a/3 = 8

process:

a/3 = 8 - 2

a/3 = 6

a = 6*3

a = 18

Check:

2 + 18/3 = 8

2 + 6 = 8

3 0

1 year ago

Csc2theta=csctheta/2costheta Can you help verify the identity

Yakvenalex [24]

csc(2x) = csc(x)/(2cos(x))

1/(sin(2x)) = csc(x)/(2cos(x))

1/(2*sin(x)*cos(x)) = csc(x)/(2cos(x))

(1/sin(x))*1/(2*cos(x)) = csc(x)/(2cos(x))

csc(x)*1/(2*cos(x)) = csc(x)/(2cos(x))

csc(x)/(2*cos(x)) = csc(x)/(2cos(x))

The identity is confirmed. Notice how I only altered the left hand side (LHS) keeping the right hand side (RHS) the same each time.

7 0

2 years ago

Find the Area of the figure below, composed of a rectangle and a semicircle.

Mamont248 [21]

Answer:

68.1 units²

Step-by-step explanation:

Start by finding the area of the rectangle...

$Area=lw\\Area=6*9\\Area=54$

Then, find the area of the semicircle...

$Radius=\frac{1}{2}*Diameter=\frac{1}{2}*6=3$

$Area=\frac{1}{2}\pi r^2\\Area=\frac{1}{2}\pi 3^2\\Area=\frac{1}{2}\pi 9\\Area=\frac{9}{2}\ pi\\Area\approx14.1$

Now, add the two to get...

54+14.1=68.1

3 0

2 years ago