All categories

3 years ago

The following figures GUNS and RUNS are parallelograms . Find x and y . (Lengths are in cm)

Mathematics

1 answer:

Bingel [31]3 years ago

4 0

Answer: (6, 9), (3, 13)

Step-by-step explanation:

Left picture:

opposite side of parallelograms are congruent (equal)

3x = 18 3y - 1 = 26

÷3 ÷3 +1 +1

x = 6 3y = 27

÷3 ÷3

y = 9

**************************************************************************

Right picture:

diagonals of a parallelogram bisect each other so the left side and right sides of each diagonal are congruent

20 = y + 7

-7 -7

13 = y

16 = x + y

16 = x + 13

-13 -13

3 = x

You might be interested in

Lease someone help me im so confused on how to find it.

LiRa [457]

The answer is D you simply get one point and move it 4 squares down and 6 squares to the right

5 0

3 years ago

Is the following relation a function: {(3,2), (3,-2),(1,-4),(-1,2)}

frez [133]

No as the x value 3 is repeated.

3 0

2 years ago

Write the given logarithmic equation in exponential form. In 4x = 9.6

KengaRu [80]

90% of people marry there 7th grade love. since u have read this, u will be told good news tonight. if u don't pass this on nine comments your worst week starts now this isn't fake. apparently if u copy and paste this on ten comments in the next ten minutes you will have the best day of your life tomorrow. you will either get kissed or asked out in the next 53 minutes someone will say i love you

7 0

3 years ago

Find the exact value of cos(sin^-1(-5/13))

son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

$\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}$

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}$

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

8 0

3 years ago

If Pam can install 12 insulators in 15 hours, how long does it take to install one insulator

Xelga [282]

The time it takes to install 1 insulator would be 1.25 hours

7 0

3 years ago

Read 2 more answers