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3 years ago

(worth 10 points) 2x + 5 = 7x - 15

Mathematics

2 answers:

Karolina [17]3 years ago

6 0

Answer:

The solution for the equation is $x = 4$ .

Step-by-step explanation:

-Solve the equation:

$2x +5=7x -15$

-Subtract 7x from 2x:

$2x +5 -7x=7x -7x -15$

$-5x + 5=-15$

-Subtract both sides by 5:

$2x + 5 - 5 = -15 -5$

$-5x = -20$

-Divide both sides by -5:

$\frac{-5x}{-5} = \frac{-20}{-5}$

$x = 4$

So, the answer is $x = 4$ .

Kaylis [27]3 years ago

5 0

The answer to the equation is x=4

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5 bullets in a 6 bullet barrell, what are your chances of dying on the first trigger pull?

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2 years ago

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A rocket is launched from a tower. The height of the rocket, y in feet, is related to the

sladkih [1.3K]

Answer:

10.17 seconds

Step-by-step explanation:

substitute y=0 into the equation

0=-16x²+153x+98

then use the quadratic formula

a = -16

b= 153

c = 98

Substituting values into the Quadratic equation, you get (Round the values to the nearest hundreth):

x₁= 10.17

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5 0

2 years ago

what is the answer here 90 student went to the zoo, 3 had hamburger milk and cake; had 5 milk and hamburger; 10 had cake and mil

dangina [55]

Answer:

a) 37

b) 2

c) 17

d) 8

Step-by-step explanation:

90 Students went to the zoo. 3 had hamburger, milk and cake; 5 had milk and hamburger, 10 had cake and milk; 8 had cake and hamburger; 24 had hamburger; 38 had cake; 20 had milk. How many had a. nothing b. cake only c. milk only d. hamburger only

Solution:

Let h represent students that ate hamburger, m represent students that had milk and c represent students that had cake.

Given that:

n(h ∩ m ∩ c) = 3, n(m ∩ h) = 5, n(c ∩ m) = 10, n(c ∩ h) = 8, n(h) = 24, n(c) = 38, n(m) = 20

The number of students that had nothing = n(h ∪ m ∪ C)'

The number of students that had only milk = n(m ∩ h' ∩ C')

The number of students that had only cake = n(m' ∩ h' ∩ C)

The number of students that had only hamburger = n(m' ∩ h ∩ C')

a) n(m ∩ h' ∩ C') = n(m) - n(m ∩ h) - n(c ∩ m) - n(h ∩ m ∩ c) = 20 - 5 - 10 - 3 = 2

n(m' ∩ h ∩ C') = n(h) - n(m ∩ h) - n(c ∩ h) - n(h ∩ m ∩ c) = 24 - 5 - 8 - 3 = 8

n(m' ∩ h' ∩ C) = n(m) - n(m ∩ c) - n(c ∩ h) - n(h ∩ m ∩ c) = 38 - 10 - 8 - 3 = 17

n(m ∩ h' ∩ C') + n(m' ∩ h ∩ C') + n(m' ∩ h' ∩ C) + n(h ∪ m ∪ C)' + n(h ∩ m ∩ c) + n(m ∩ h) + n(c ∩ m) + n(c ∩ h) = 90

2 + 8 + 17 + 5 + 10 + 8 + 3 + n(h ∪ m ∪ C)' = 90

53 + n(h ∪ m ∪ C)' = 90

n(h ∪ m ∪ C)' = 37

b) n(m' ∩ h' ∩ C) = 17

c) n(m ∩ h' ∩ C') = 2

d) n(m' ∩ h ∩ C') = 8

4 0

2 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

Help please. I need to finish this in five minutes!

e-lub [12.9K]

Answer:

i have no idea....

Step-by-step explanation:

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3 0

3 years ago