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3 years ago

Determine if the two triangles are congruent. If they are, State how you know. NO LINKS!!!! Show your work. Part 2c

Mathematics

1 answer:

tankabanditka [31]3 years ago

7 0

Answer:

2. Not enough information

4. Congruent SAS

4. Similar, not enough information to determine congruency.

Step-by-step explanation:

2. We only know one side and one angle are congruent, Not enough to determine congruency

4. We know two sides and the angle between are vertical angles and vertical angles are congruent. SAS is how the triangles are congruent.

6. The three angles are congruent which makes the triangles similar. We need to know a side if they are to be congruent

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Elis [28]

768 because 6x8 = 48 the 8 drop Down to the answer and the 4 on top of 1 and multiply 1x8= 8+1 =9 so the 9 drop down to the answer. Now that is fake because you need to multiply 4x6 and 4x1 so 4x6 is 24 below the 8 add a 0 and go 1 move to the left and put 4 and 2 on top of 1 now 4x1 is 4+1 is 5 and put 5 on the left of 4 and the final step is add 98+540=768 so 768 is your real answer.

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3 years ago

E.) A ball is thrown into the air with an initial upward

Furkat [3]

Answer:

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Step-by-step explanation:

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A random sample of 64 freshmen spent an average of 14 hours per week watching television. the sample standard deviation was 32 h

NemiM [27]

Answer:

+/- 1.9983

Step-by-step explanation:

The critical values for a confidence interval for the population mean, with population standard deviation not known, are obtained from the student's t distribution.

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100 - 95 = 5%.

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+/- 1.9983

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3 years ago

Which is an equation of the line passing

tia_tia [17]

$\bf (\stackrel{x_1}{10}~,~\stackrel{y_1}{8})~\hspace{10em} slope = m\implies \cfrac{2}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-8=\cfrac{2}{5}(x-10) \\\\\\ y-8=\cfrac{2}{5}x-4\implies y=\cfrac{2}{5}x+4$

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3 years ago

Integrate the following problem:

vazorg [7]

Answer:

$\displaystyle \frac{2 \cdot sin2x-cos2x}{5e^x} + C$

Step-by-step explanation:

The integration by parts formula is: $\displaystyle \int udv = uv - \int vdu$

Let's find u, du, dv, and v for $\displaystyle \int e^-^x \cdot cos2x \ dx$ .

$u=e^-^x$
$du=-e^-^x dx$

$dv=cos2x \ dx$
$v= \frac{sin2x}{2}$

Plug these values into the IBP formula:

$\displaystyle \int e^-^x \cdot cos2x \ dx = e^-^x \cdot \frac{sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$
$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$

Now let's evaluate the integral $\displaystyle \int \frac{sin2x}{2} \cdot -e^-^x dx$ .

Let's find u, du, dv, and v for this integral:

$u=-e^-^x$
$du=e^-^x dx$
$dv=\frac{sin2x}{2} dx$
$v=\frac{-cos2x}{4}$

Plug these values into the IBP formula:

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} - \int \frac{-cos2x}{4}\cdot e^-^x dx$

Factor 1/4 out of the integral and we are left with the exact same integral from the question.

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx$

Let's substitute this back into the first IBP equation.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Simplify inside the brackets.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ \frac{e^-^x \cdot cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Distribute the negative sign into the parentheses.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4} - \frac{1}{4} \int cos2x \cdot e^-^x dx$

Add the like term to the left side.

$\displaystyle \int e^-^x \cdot cos2x \ dx + \frac{1}{4} \int cos2x \cdot e^-^x dx= \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$
$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$

Make the fractions have common denominators.

$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x}{4} - \frac{e^-^x \cdot cos2x}{4}$

Simplify this equation.

$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4}$

Multiply the right side by the reciprocal of 5/4.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4} \cdot \frac{4}{5}$

The 4's cancel out and we are left with:

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{5}$

Factor $e^-^x$ out of the numerator.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x(2 \cdot sin2x-cos2x)}{5}$

Simplify this by using exponential properties.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x}$

The final answer is $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x} + C$ .

7 0

3 years ago

Read 2 more answers

Determine if the two triangles are congruent. If they are, State how you know. NO LINKS!!!! Show your work. Part 2c​

Determine if the two triangles are congruent. If they are, State how you know. NO LINKS!!!! Show your work. Part 2c