Ask question

Ask question

All categories

4 years ago

13

Is it possible for two vertical angles to also be complementary angles? Explain why or why not. If so, find the measure of these

angles.

1 answer:

enot [183]4 years ago

8 0

Vertical angles are congruent. Complementary angles add to 90 degrees. So if the vertical angles are each half, 45 degrees, they'll be congruent and complementary.

Answer: 45 degrees

You might be interested in

(x^2 - x^(1/2))/(1-x^(1/2))

Levart [38]

$\frac { \left( { x }^{ 2 }-{ x }^{ \frac { 1 }{ 2 } } \right) }{ \left( 1-{ x }^{ \frac { 1 }{ 2 } } \right) }$

$\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x } \right) }{ \left( 1-\sqrt { x } \right) } \cdot 1$

$\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x } \right) }{ \left( 1-\sqrt { x } \right) } \cdot \frac { \left( 1+\sqrt { x } \right) }{ \left( 1+\sqrt { x } \right) }$

$\\ \\ =\frac { { x }^{ 2 }+{ x }^{ 2 }\sqrt { x } -\sqrt { x } -x }{ 1+\sqrt { x } -\sqrt { x } -x }$

$\\ \\ =\frac { -\sqrt { x } \left( 1-{ x }^{ 2 } \right) -x\left( 1-x \right) }{ \left( 1-x \right) }$

$\\ \\ =\frac { -\sqrt { x } \left( 1+x \right) \left( 1-x \right) -x\left( 1-x \right) }{ \left( 1-x \right) }$

$\\ \\ =\frac { \left( 1-x \right) \left\{ -\sqrt { x } \left( 1+x \right) -x \right\} }{ \left( 1-x \right) }$

$\\ \\ =-\sqrt { x } \left( 1+x \right) -x\\ \\ =-{ x }^{ \frac { 1 }{ 2 } }\left( 1+{ x }^{ \frac { 2 }{ 2 } } \right) -x$

$\\ \\ =-{ x }^{ \frac { 1 }{ 2 } }-{ x }^{ \frac { 3 }{ 2 } }-x\\ \\ =-\sqrt { x } -\sqrt { { x }^{ 3 } } -x$

3 0

3 years ago

Read 2 more answers

3.8:(4c+3)=2:2 2/19<br><br>Please help me! I really appreciate it!

Sati [7]

Please note that in such questions, the ":" stands for ratio which in turn stands for divide. Thus, the given expression which is:

$3.8:(4c+3)=2:2\frac{2}{19}$ will become:

$\frac{3.8}{4c+3}= \frac{2}{2\frac{2}{19}}$ .....(Equation 1)

Now, we know that 3.8 can be rewritten as $\frac{38}{10}$ . Also, we know that the mixed fraction $2\frac{2}{19}$ can be re-written as:

$2\frac{2}{19}=2+ \frac{2}{19}= \frac{38+2}{19}= \frac{40}{19}$

Using the above two expressions in (Equation 1) we get:

$\frac{\frac{38}{10}}{4c+3}= \frac{2}{\frac{40}{19}}$

$\frac{\frac{38}{10}} {4c+3}= \frac{2\times 19}{40}= \frac{38}{40}$

Cross-multiplication gives:

$38(4c+3)=\frac{40\times 38}{10}$

$4c+3=4$ $4c=1$

$c=\frac{1}{4}$

Or c=0.25

Thus, we may write the value of c as either the decimal 0.25 or the fraction $\frac{1}{4}$ .

8 0

3 years ago

Read 2 more answers

The highway department is testing two types of reflecting paint for concrete bridge end pillars. The two kinds of paint are alik

tankabanditka [31]

Answer:

a. difference of means

Step-by-step explanation:

Given that :

Mean , x = 9.4

Standard deviation, $s.d_1$ = 2.5

Number, $n_1$ = 12

Mean, y = 6.5

standard deviation, $s.d_2$ = 2.4

Number, $n_2$ = 12

The null hypothesis is : $H_0: \mu_1=\mu_2$

The alternate hypothesis is : $H_1: \mu_1>\mu_2$

Level of significance, $\alpha$ = 0.1

From the $\text{standard normal table, right tailed,}$ $t_{1/2}$ = 1.363

Since out test is right tailed.

Reject $H_0$ , if $T_0>1.363$

We use the test statics,

$t_0=\frac{(x-y)}{\sqrt{\frac{s.d_1}{n_1}+\frac{s.d_2}{n_2}}}$

$t_0=\frac{(9.4-6.5)}{\sqrt{\frac{6.25}{12}+\frac{5.76}{12}}}$

$t_0=2.899$

$|t_0|=2.899$

$\text{Critical value}$

The value of $|t_{1/2}|$ with minimum $\left(n_1-1,n_2-1)$ that is 11 df is 1.363

We go $|t_0|=2.899$ and $|t_{1/2}|$ = 1.363

Decision making:

Since the value of $|t_0|>|t_{1/2}|$$ and we reject the $H_0$

The p-value : right tail $H_a:(p>2.8988)$

= 0.00724

Therefore the value of $p_{0.1} > 0.00724$ , and so we reject the $H_0$

Thus we are testing 'the difference of means" in this problem.

7 0

3 years ago

What Values Will be in the table on the right ?

Readme [11.4K]

Jsjsjsjjsjsjsjns s s nsnsjsjjdjs

3 0

3 years ago

How many vertices does a closed cone have?

Allushta [10]

Answer:

A closed cone has 0 verticals.. but 1 vertice

Step-by-step explanation:

5 0

3 years ago