Is the decimal form of 13/3 a rational number?

5/2b+(-25/2)=-10 what is the value of b?

gulaghasi [49]

The value of b in the equation 5/2b+(-25/2) is 1.

8 0

3 years ago

Right triangles abc and dbc with right angle c are given below. If cos(a)=15,ab=12 and cd=2, find the length of bd.

dmitriy555 [2]

see the attached figure to better understand the problem

we have that

$cos(A)=\frac{1}{5} \\ AB=12\ units\\ CD=2\ units$

Step 1

<u>Find the value of AC</u>

we know that

in the right triangle ABC

$cos (A)=(AC/AB)\\AC=AB*cos(A)$

substitute the values in the formula

$AC=12*(1/5)\\ AC=2.4\ units$

Step 2

<u>Find the value of BC</u>

we know that

in the right triangle ABC

Applying the Pythagorean Theorem

$AB^{2} =AC^{2}+BC^{2}\\ BC^{2}=AB^{2} -AC^{2}$

substitute the values

$BC^{2}=12^{2} -2.4^{2}\\BC^{2}= 138.24\\ BC=11.76\ units$

Step 3

<u>Find the value of BD</u>

we know that

in the right triangle BCD

Applying the Pythagorean Theorem

$BD^{2} =DC^{2}+BC^{2}$

substitute the values

$BD^{2} =2^{2}+11.76^{2}$

$BD=11.93\ units$

therefore

<u>the answer is</u>

the length of BD is 11.93 units

8 0

3 years ago

Read 2 more answers

Can someone please help me with this question?

Vanyuwa [196]

Answer:

5 seconds

Step-by-step explanation:

In 5 seconds it will be at 98 meters

x = -b/2a

x = -20/2(-2) = -20/-4 = 5

8 0

2 years ago

In triangles ABC and LMN, ∠A ≅ ∠L, ∠B ≅ ∠M, and ∠C ≅ ∠N. Is this information sufficient to prove triangles ABC and LMN congruent

ipn [44]

Answer:

No

Step-by-step explanation:

No. We have 3 angles of triangle ABC congruent to 3 corresponding angles of triangle LMN. That is AAA. To use ASA, you need two angles and an included side. With three angles, you can prove the triangles similar, but not congruent.

5 0

2 years ago

Evaluate the following limit:

Makovka662 [10]

If we evaluate the function at infinity, we can immediately see that:

$\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle L = \lim_{x \to \infty}{\frac{(x^2 + 1)^2 - 3x^2 + 3}{x^3 - 5}} = \frac{\infty}{\infty}} \end{gathered}$}$

Therefore, we must perform an algebraic manipulation in order to get rid of the indeterminacy.

We can solve this limit in two ways.

By comparison of infinities:

We first expand the binomial squared, so we get

$\large\displaystyle\text{$\begin{gathered}\sf \displaystyle L = \lim_{x \to \infty}{\frac{x^4 - x^2 + 4}{x^3 - 5}} = \infty \end{gathered}$}$

Note that in the numerator we get x⁴ while in the denominator we get x³ as the highest degree terms. Therefore, the degree of the numerator is greater and the limit will be \infty. Recall that when the degree of the numerator is greater, then the limit is \infty if the terms of greater degree have the same sign.

Dividing numerator and denominator by the term of highest degree:

$\large\displaystyle\text{$\begin{gathered}\sf L = \lim_{x \to \infty}\frac{x^{4}-x^{2} +4 }{x^{3}-5 } \end{gathered}$}\\$

$\ \ = \lim_{x \to \infty\frac{\frac{x^{4} }{x^{4} }-\frac{x^{2} }{x^{4}}+\frac{4}{x^{4} } }{\frac{x^{3} }{x^{4}}-\frac{5}{x^{4}} } }$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{=\lim_{x \to \infty}\frac{1-\frac{1}{x^{2} } +\frac{4}{x^{4} } }{\frac{1}{x}-\frac{5}{x^{4} } } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{0}=\infty } \end{gathered}$}$

Note that, in general, 1/0 is an indeterminate form. However, we are computing a limit when x →∞, and both the numerator and denominator are positive as x grows, so we can conclude that the limit will be ∞.

5 0

2 years ago