All categories

3 years ago

Can you solve 2x+16 = 40

Mathematics

2 answers:

EastWind [94]3 years ago

6 0

Answer:

Step-by-step explanation:

work backwards to find x. you start with 40, then subtract 16. now you have 2x= 24. now you have to figure out what multiplied by 2 is 24. 24 divided by 2 is 12, there is the answer. x=12. Always check your work, solve it normally replacing x with 12.

Vera_Pavlovna [14]3 years ago

4 0

Answer:x=12

Step-by-step explanation:subtract 16 from both sides (16-16=0) (40-16=24)

2x=24 divide 2 to both sides (2/2=0) (24/2=12) hope i explained it correctly.

you always need to get x by itself so whatever is in from of x divide that by what ever is behind the equal(=) sign.

You might be interested in

I will give your brainliest for this please

VashaNatasha [74]

<h2>Answer: Where is the question</h2>

6 0

2 years ago

Prove the quadratic formula

STALIN [3.7K]

ax² + bx + c = 0

x = (-b ± √(b² - 4ac))/2a

First, rewrite the first equation so that the first coefficient is 1. Divide everything by a.

(ax² + bx + c = 0)/a =

x² + (b/a)x + (c/a) = 0

Isolate (c/a) by subtracting (c/a) from both sides

x² + (b/a)x + (c/a) (-(c/a) = 0 (- (c/a)

x² + (b/a)x = 0 - (c/a)

Add spaces

x² + (b/a)x = -c/a

Take 1/2 of the middle term's coefficient and square it. Remember that what you add to one side, you add to the other.

x² + (b/a)x + (b/2a)² = -c/a + (b/2a)²

Simplify the left side of the equation.

x² + (b/a)x + (b/2a)² = (x + (b/2a))²

(x + b/2a))² = ((b²/4a²) - (4ac/4a²)) -> ((b² - 4ac)/(4a²))

Take the square root of both sides of the equation

√(x + b/2a))² = √((b²/4a²) - (4ac/4a²))

x + b/(2a) = (±√(b² - 4ac)/2a

Simplify. Isolate the x.

x = -(b/2a) ± (∛b² - 4ac)/2a = (-b ± √(b² - 4ac))/2a

3 0

3 years ago

Which of these relations on the set {0, 1, 2, 3} are equivalence relations? If not, please give reasons why. (In other words, if

Delicious77 [7]

Answer:

(1)Equivalence Relation

(2)Not Transitive, (0,3) is missing

(3)Equivalence Relation

(4)Not symmetric and Not Transitive, (2,1) is not in the set

Step-by-step explanation:

A set is said to be an equivalence relation if it satisfies the following conditions:

Reflexivity: If $\forall x \in A, x \rightarrow x$
Symmetry: $\forall x,y \in A, $if x \rightarrow y,$ then y \rightarrow x$
Transitivity: $\forall x,y,z \in A, $if x \rightarrow y,$ and y \rightarrow z, $ then x \rightarrow z$

(1) {(0,0), (1,1), (2,2), (3,3)}

(3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)}

The relations in 1 and 3 are Reflexive, Symmetric and Transitive. Therefore (1) and (3) are equivalence relation.

(2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)}

In (2), (0,2) and (2,3) are in the set but (0,3) is not in the set.

Therefore, It is not transitive.

As a result, the set (2) is not an equivalence relation.

(4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}

(1,2) is in the set but (2,1) is not in the set, therefore it is not symmetric

Also, (2,0) and (0,1) is in the set, but (2,1) is not, rendering the condition for transitivity invalid.

5 0

3 years ago

Problem

Sergeu [11.5K]

The lengths of HJ and JK are 50 and 72 respectively.

Given: J is between H and K. the length of HJ=2x+4, JK=3x+3, and HK=22.

The line HJ and JK are collinear then the total length is given as: HJ+JK=HK

substitute values

⇒ (2x+4)+(3x+3)=22

⇒ 5x+7=22

⇒ 5x=22-7

⇒ 5x=15

⇒ x=3

Thus, the length of HJ is (2x+4)=(2*3+4)=10

And the length of JK is (3x+3)=(3*3+3)=12

Learn more about collinear lines here: brainly.com/question/27943430

#SPJ10

8 0

2 years ago

Read 2 more answers

How do I get the answer y= 2x^2+9x-5

guapka [62]

Answer:

Quadratic Formula

x = -5

and

x = 0.5

Step-by-step explanation:

Whenever you see a problem in this form, which you will see a lot of, you can try to factor it or use the "least squares" method or what have you, but those won't always work, unfortunately.

Fortunately, the quadratic formula will never fail you with quadratic expressions.

This is the Quadratic Formula

$x = \frac{-b \frac{+}{-} \sqrt{b^2-4ac} }{2a}$

a is the the number on the variable with the exponent ^2

b is the number on the variable with no exponent

c is the third number

a and b cannot be equal to 0; c can be

Since we're looking for a number with an equation that has a square root in it, we're going to get two answers. These two answers come from the radical being separately added AND subtracted from the radical. It's basically two problems.

Plugging in our numbers to this equation gives us x values of -5 and 0.5. This will always work with polynomials with factors of ^2 in them.

If you have a TI-84 calculator or newer, there's a tool on it that will factor polynomials like this one for you just by giving it the numbers.

7 0

3 years ago