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2 years ago

Please solve! It's not A.

Mathematics

2 answers:

bixtya [17]2 years ago

7 0

Answer:

line 2?

Step-by-step explanation:

yawa3891 [41]2 years ago

3 0

It’s line 2 I took the test I found the answer in quizlet

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ANSWER ASAP1! GET BRAINLEST

Oksanka [162]

Answer:

x = $2\dfrac{3}{7}$

Explanation:

$\hookrightarrow 35.1 = 12.9 + 10.2x$

$\hookrightarrow 10.2x = 35.1 - 12.9$

$\hookrightarrow \dboxed{10.2x = 22.2}$

$\hookrightarrow x =\dfrac{22.2}{10.2}$ * 5

$\hookrightarrow x = \dfrac{111}{51}$

$\hookrightarrow x = \dfrac{37}{17}$

$\hookrightarrow x = 2\dfrac{3}{7}$

3 0

3 years ago

Read 2 more answers

Please help !!! I’ll mark you as brainliest if correct

kozerog [31]

Answer:

10 seconds

Step-by-step explanation:

we are given the height (s) which is 1600ft. and the formula—where t is time in seconds— t= 1/4 $\sqrt{s}$ , so we can plug 1600ft in for s and solve:

t= 1/4 $\sqrt{1600}$

t= 1/4(40)

t= 10 seconds

5 0

3 years ago

Difference between greatest and least possible areas of 18 in. Perimeter

Kipish [7]

Least: 8 (length) x 1(width) = 8
Greatest: 5 (length x 4 (width) = 20
The difference between them: 20 - 8 = 12<span />

6 0

3 years ago

Find Horizontal and Vertical Asymptotes for y=(1)/((x+2)^3)

Usimov [2.4K]

Answer:

Vertical asymptote: $x = -2$

Horizontal asymptote: $y = 0$ or x axis.

Step-by-step explanation:

The rational function is given as:

$y=\frac{1}{(x+2)^3}$

Vertical asymptotes are those values of $x$ for which the function is undefined or the graph moves towards infinity.

For a rational function, the vertical asymptotes can be determined by equating the denominator equal to zero and finding the values of $x$ .

Here, the denominator is $(x+2)^3$

Setting the denominator equal to zero, we get

$(x+2)^3=0\\(x+2)=0\\x=-2$

Therefore, the vertical asymptote occur at $x=-2$ .

Horizontal asymptotes are the horizontal lines when $x$ tends towards infinity.

For a rational function, if the degree of numerator is less than that of the denominator, then the horizontal asymptote is given as $y=0$ .

Here, there is no $x$ term in the numerator. So, degree is 0. The degree of the denominator is 3. So, the degree of numerator is less than that of denominator.

Therefore, the horizontal asymptote is at $y=0$ or x axis.

3 0

3 years ago

Need help hard question! Picture has it all.

iris [78.8K]

Using the tagent rule:

Tangent(angle) = Opposite Leg / Adjacent leg

Tangent(51) = x / 10

x = 10 * Tangent (51)

x = 12.349

x = 12.3 ( Rounded to the nearest tenth)

3 0

3 years ago