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3 years ago

Find the size of unknown angles

Mathematics

1 answer:

Luden [163]3 years ago

4 0

Answer:

ok in this diagram we have two triangles the first triangle all angles are unknown the second triangle we have the values of two angles so we will start with the second triangle and as you know angles in a triangle add up to 180° so we will take 35 degrees + 30 degrees and name the unknown angle x is equals to 180 degrees so it will be 65 degrees + x is equals to 180° put the like terms together x will be equal to 180 degrees - 65 degrees it becomes a minus because when a plus crosses the equal sign it becomes - so x is equals to 115° and as you can see in the diagram 115° is vertically opposite z then you can extend the lines so as to understand the diagram better when you extend the lines you find that 35 degrees is equal to angle v so it will be 115 degrees + 35 degrees + x is equals to 180 degrees so 150° + x is equals to 180 degrees put the like times together x is equals to 180 degrees minus 150° x is equals to 30 degrees

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Find the slope of the line below using rise/run.<br>

QveST [7]

Answer:

A, 2/3

Step-by-step explanation:

you rise 2 and run 3

6 0

3 years ago

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PLEASE ANSWER + BRAINLIEST !!!!<br><br> Find the indicated sum for the arithmetic series.

schepotkina [342]

Hello there,

I believe the answer would be 17×6 - 6(9+14)(6)/2 = 102 -414 = -312 but i am not 100% Sure!! Sorry if I am wrong!

Hope I helped

8 0

3 years ago

Fill in Sin, Cos, and tan ratio for angle x. <br> Sin X = 4/5 (28/35 simplified)

Fantom [35]

Answer:

Given: $\sin(x) = (4/5)$ .

Assuming that $0 < x < 90^{\circ}$ , $\cos(x) = (3/5)$ while $\tan(x) = (4/3)$ .

Step-by-step explanation:

By the Pythagorean identity $\sin^{2}(x) + \cos^{2}(x) = 1$ .

Assuming that $0 < x < 90^{\circ}$ , $0 < \cos(x) < 1$ .

Rearrange the Pythagorean identity to find an expression for $\cos(x)$ .

$\cos^{2}(x) = 1 - \sin^{2}(x)$ .

Given that $0 < \cos(x) < 1$ :

$\begin{aligned} &\cos(x) \\ &= \sqrt{1 - \sin^{2}(x)} \\ &= \sqrt{1 - \left(\frac{4}{5}\right)^{2}} \\ &= \sqrt{1 - \frac{16}{25}} \\ &= \frac{3}{5}\end{aligned}$ .

Hence, $\tan(x)$ would be:

$\begin{aligned}& \tan(x) \\ &= \frac{\sin(x)}{\cos(x)} \\ &= \frac{(4/5)}{(3/5)} \\ &= \frac{4}{3}\end{aligned}$ .

7 0

2 years ago

Look at the graph shown below:

ElenaW [278]

So.. take a peek at the picture... let's get two points from it, hmm say 0,4 notice it touches the y-axis there, and say hmmm -4, 1, almost at the bottom of the line

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 0}}\quad ,&{{ 4}})\quad 
% (c,d)
&({{ -4}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-4}{-4-0}$

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad 
\begin{array}{llll}
\textit{plug in the values for }
\begin{cases}
y_1=4\\
x_1=0\\
m=\boxed{?}
\end{cases}\\
\textit{and solve for "y"}
\end{array}\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}$

once you get the slope and solve for "y", that'd be the equation of the line.

7 0

3 years ago

If f(1)=10 and f(n)=-5f(n-1)-2 then find the value of f(4)

Flura [38]

Answer:

$\large\boxed{f(4)=-1292}$

Step-by-step explanation:

$f(n)=-5f(n-1)-2\\\\f(1)=10\\\\f(2)=-5f(2-1)-2=-5f(1)-2\\\text{put}\ f(1)=10:\\f(2)=-5(10)-2=-50-2=-52\\\\f(3)=-5f(3-1)-2=-5f(2)-2\\\text{put}\ f(2)=-52\\f(3)=-5(-52)-2=260-2=258\\\\f(4)=-5f(4-1)-2=-5f(3)-2\\\text{put}\ f(3)=258\\f(4)=-5(258)-2=-1290-2=-1292$

4 0

3 years ago