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3 years ago

13

Negative two thirds • one and one third

1 answer:

Anuta_ua [19.1K]3 years ago

4 0

Answer:

-8/9 or -0.8 repeating

Step-by-step explanation:

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X + x + x = 54<br> What’s the answer for this

Oliga [24]

Answer:

x = 18

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

x+x+x=54

(x+x+x)=54(Combine Like Terms)

3x=54

3x=54

Step 2: Divide both sides by 3.

3x 3 = 54

3 x=18

Answer:

x=18

8 0

3 years ago

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Estimate the distance you can travel in 3 hours 20 minutes if you drive on average 49 miles per hour. Round your answer to the n

AlexFokin [52]

Answer:

163 miles

Step-by-step explanation:

set up a ratio based on 49 miles in 1 hour (60 minutes):

let 'd' = distance in 3hrs, 20 min

3hrs, 20 min = 200 minutes

49/60 = d/200

cross-multiply:

60d = 9800

d = 163.3

7 0

2 years ago

Tia performed an experiment where she flipped a coin 200 times. The coin landed heads up 92 times and tails up 108 times. Which

11111nata11111 [884]

only the first statement is true - it is the experimental probability. the rest is incorrect: the ratio is not the number of trials; the theoretical probability should be 0.5 (for unbiased coins); ratio never represents a number of occurences.

4 0

4 years ago

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Anyone have an idea?

Phantasy [73]

Answer:

yes I have an idea jnhvhbvcf

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4 years ago

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Х- а<br>x-b<br>If f(x) = b.x-a÷b-a + a.x-b÷a - b<br>Prove that: f (a) + f(b) = f (a + b)

GenaCL600 [577]

Given:

Consider the given function:

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot(x-b)}{a-b}$

To prove:

$f(a)+f(b)=f(a+b)$

Solution:

We have,

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot (x-b)}{a-b}$

Substituting $x=a$ , we get

$f(a)=\dfrac{b\cdot(a-a)}{b-a}+\dfrac{a\cdot (a-b)}{a-b}$

$f(a)=\dfrac{b\cdot 0}{b-a}+\dfrac{a}{1}$

$f(a)=0+a$

$f(a)=a$

Substituting $x=b$ , we get

$f(b)=\dfrac{b\cdot(b-a)}{b-a}+\dfrac{a\cdot (b-b)}{a-b}$

$f(b)=\dfrac{b}{1}+\dfrac{a\cdot 0}{a-b}$

$f(b)=b+0$

$f(b)=b$

Substituting $x=a+b$ , we get

$f(a+b)=\dfrac{b\cdot(a+b-a)}{b-a}+\dfrac{a\cdot (a+b-b)}{a-b}$

$f(a+b)=\dfrac{b\cdot (b)}{b-a}+\dfrac{a\cdot (a)}{-(b-a)}$

$f(a+b)=\dfrac{b^2}{b-a}-\dfrac{a^2}{b-a}$

$f(a+b)=\dfrac{b^2-a^2}{b-a}$

Using the algebraic formula, we get

$f(a+b)=\dfrac{(b-a)(b+a)}{b-a}$ $[\because b^2-a^2=(b-a)(b+a)]$

$f(a+b)=b+a$

$f(a+b)=a+b$ [Commutative property of addition]

Now,

$LHS=f(a)+f(b)$

$LHS=a+b$

$LHS=f(a+b)$

$LHS=RHS$

Hence proved.

5 0

3 years ago