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3 years ago

Find the value of x to the nearest tenth.

Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

8 0

Answer:

The value of x = 4√5

Step-by-step explanation:

<u>Points to remember</u>

For a right angled triangle

Hypotenuse² = Base² + Height²

<u>To find the value of x</u>

From the figure we can see a right angled triangle with,

hypotenuse = 6 and height = 4

Value of x = 2 * base

we have, Hypotenuse² = + Height²

Base² = Hypotenuse² - Height²

= 6² - 4²

= 36 - 16 = 20

Base = √20 = 2√5

x = 2 * 2√5 = 4√5

The value of x = 4√5

Kryger [21]3 years ago

5 0

Answer:

x = 8.9 (nearest tenth)

Step-by-step explanation:

6^2 - 4^2 = 36 - 16 = 20

x = 2 * (√20)

x = 2 * 4.47

x = 8.94

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Y=5(3^x)+4 identity the y intercept

Kisachek [45]

Y - intercept =(0,9)

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3 years ago

Which of the following is not a solution for the inequality 3x +2 < 12?

DIA [1.3K]

Answer:

x = 3.5 is not a solution

Step-by-step explanation:

3x +2 < 12

Subtract 2 from each side

3x +2-2 < 12-2

3x< 10

Divide each side by 3

3x/3 <10/3

x < 10/3

x <3.3333333

Any number greater than 3.3333 is not a solutions

3.5 is not a solution

5 0

3 years ago

Read 2 more answers

"assume that s is a non-empty set of real numbers which is bounded above and lambda is the least upper bound. Prove that for all

s2008m [1.1K]

By definition, if $\lambda$ is the least upper bound of the set $S$ , it means two thing:

$\forall x \in S,\ x \leq \lambda$
$\forall \varepsilon>0,\ \exists x \in S:\ x>\lambda-\varepsilon$

In other words, the least upper bound of a set is greater than or equal to every single element of the set, but it is "close enough" to the elements of the set, because you guaranteed to find elements in the set between $\lambda-\varepsilon$ and $\lambda$

For example, pick $S = [1,10)$ . Obvisouly, the least upper bound is $\lambda = 10$ . In fact, every number in $[1,10)$ is smaller than 10, but as soon as you take away something from 10, say 0.01, you get 9.99, and there are elements in $S$ greater than 9.99, say 9.9999.

So, the claim is basically proven by definition: if $\beta < \lambda$ , let $0 < \delta = \lambda - \beta$ . By definition, there exists $\alpha \in S:\ \alpha > \lambda - \delta$ .