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3 years ago

A triangle has side lengths of 14 cm and 8 cm. Which could be the value of the third side, 26 cm or 18 cm?

1 answer:

8 0

It’s 18cm because to get the lengths of the triangle you have to add 14+8 and the answer has to be greater than the 3rd side. When I added 14+8, I got 22 and 22 is greater than 18. So 18cm is the answer.
Hope this helps!

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(MULTIPLE CHOICE QUESTION)

weeeeeb [17]

Answer:

Transitive property is used in step 4.

Step-by-step explanation:

Transitive property states that if

$a = b \\b = c\\ then\\a = c$

Here we have

$\angle 5 \cong \angle 1\\\angle 1 \cong \angle 3\\then\\\angle 3 \cong \angle 5$

3 0

3 years ago

Read 2 more answers

Please help!! Due really soon!

Alisiya [41]

9514 1404 393

Answer:

f(x) = 6x +1

Step-by-step explanation:

Differences in x-values (first row) are 1, 1, 1.

Differences in y-values (second row) are 6, 6, 6.

The constant ratio of differences (6/1) tells you the function is linear, and has a slope of m = 6/1 = 6.

Using the first point in the form ...

y = mx + b

we have ...

y = 6x + b

7 = 6·1 + b . . . . (x, y) = (1, 7)

1 = b . . . . . . subtract 6

Then the equation can be written ...

y = 6x +1

In functional form, this is ...

f(x) = 6x +1

4 0

3 years ago

-8x-2y=6 in slope form

Harrizon [31]

Perhaps you mean "slope-intercept" form. Solve for y and reduce the fractions.
.. -8x -6 = 2y . . . . . . . . add 2y-6
.. y = -4x -3 . . . . . . . . . divide by 2

Your line in slope-intercept form is
.. y = -4x -3

4 0

4 years ago

Zina solves a system of linear equations by elimination and finds a solution of (2, 2). One of the equations is a + b = 4. What

Dima020 [189]

Answer:

Option D

a+3b=8

6 0

3 years ago

Read 2 more answers

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

1 year ago