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2 years ago

Help?? 15 points for whoever helps! QUICK!!

Mathematics

1 answer:

Tcecarenko [31]2 years ago

5 0

Hii.

We know that we have to do LxW,or length times width.

So, what is 7.2cm x 13cm?

93.6

So, the answer would be 93.6

I hope dis helps you.

Please,let me know if im wrong.

~Learning with Natalia~

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Use the graphing calculator to graph the quadratic function y = 2x2 – 6x – 8. Which values are solutions of 0 = 2x2 – 6x – 8? Ch

SashulF [63]

4 is your answer. hope its correct

7 0

3 years ago

Read 2 more answers

Find the composition of

Rus_ich [418]

Answer:

[(x + 6), (y + 1)]

Step-by-step explanation:

Vertices of the quadrilateral ABCD are,

A → (-5, 2)

B → (-3, 4)

C → (-2, 4)

D → (-1, 2)

By reflecting the given quadrilateral ABCD across x-axis to form the image quadrilateral A'B'C'D',

Rule for the reflection of a point across x-axis is,

(x, y) → (x , -y)

Coordinates of the image point A' will be,

A(-5, 2) → A'(-5, -2)

From the picture attached, point E is obtained by translation of point A'.

Rule for the translation of a point by h units right and k units up,

A'(x+h, y+k) → E(x', y')

By this rule,

A'(-5 + h, -2 + k) → E(1, -1)

By comparing coordinates of A' and E,

-5 + h = 1

h = 6

-2 + k = -1

k = 1

That means

Rule for the translation will be,

[(x + 6), (y + 1)]

8 0

2 years ago

A student has gotten the following grades on his tests. What is the minimum grade he must get on the last test in order to have

steposvetlana [31]

In order to get an 81 average,
the student has to get 81x5=405

71+96+90+73+?=405
330+?=405
405-330=75
he needs to get an 75

3 0

3 years ago

What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

5 0

3 years ago

1. Which of the given equations are linear?

jeka57 [31]

Which equation do you mean?

8 0

3 years ago