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Liula [17]
3 years ago
8

Find the value of |-4 3| · 8

Mathematics
1 answer:
S_A_V [24]3 years ago
6 0
<span>|-4 3| is called an absolute value, anything in an absolute value becomes a positive number. So the </span><span>|-4 3| becomes a 43. We rewrite the problem 43 * 8 = 344.</span>
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Can anyone help me?
goldfiish [28.3K]

Given:

A right angle triangle ABC, right angled at C.

To find:

The measure of θ.

Solution:

In a right angle triangle,

\tan \theta=\dfrac{Opposite}{Adjacent}

In triangle ABC,

\tan \theta=\dfrac{AC}{BC}

\tan \theta=\dfrac{2.7}{5}

\tan \theta=0.54

It can be written as

\theta=\tan^{-1}0.54

\theta\approx 28.369

Therefore, the measure of θ is 28.369.

5 0
3 years ago
Find the highest common factor of 35 and 8C.
Kaylis [27]

Answer:

Factors of 35 = 1, 5, 7 and 35

Factors of 8= 1, 2, 4, 8.

common = 1

7 0
2 years ago
A 64-ounce container of sports juice costs $6.50. A 48-ounce container of the same juice costs $4.25. Which juice container is t
larisa86 [58]

Answer:

The juice container that is a better buy is the 48 ounce container because the price per ounce is lower.

Step-by-step explanation:

In order to determine the juice container that is the better buy, you have to find the one that has a lower cost per ounce by dividing the price of each juice container by the number of ounces it has:

64-ounce container: 6.50/64=0.10

48-ounce container: 4.25/48=0.08

According to this, the answer is that the juice container that is a better buy is the 48 ounce container because the price per ounce is lower.

6 0
3 years ago
Factor completely. x2−11x+24
Serjik [45]

Step-by-step explanation:

x2-8x-3x+24=0

x(x-8) 3(x+8)

(x-8)(x+8)

x-8=0 or x+8=0

x=8. x= -8

6 0
2 years ago
For what value of constant c is the function k(x) continuous at x = 0 if k =
nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x)  is continuous, we take the limit of the parameter as follows:

\mathbf{ \lim_{x \to 0^-} k(x) =  \lim_{x \to 0^+} k(x) =  0 }

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= c }

Provided that:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }

Using l'Hospital's rule:

\mathbf{\implies  \lim_{x \to 0} \ \  \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}=  \lim_{x \to 0}   sec \ x  \ tan \ x = 0}

Therefore:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}=0 }

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

5 0
2 years ago
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