Ask question

Ask question

All categories

2 years ago

6

The diagonals of a quadrilateral bisect at right angles. What type of quadrilateral is this?

1 answer:

arsen [322]2 years ago

8 0

square

Step-by-step explanation:

You might be interested in

I need help with questions #7 and #8 plz

katen-ka-za [31]

Answer:

7. A = 40.8 deg; B = 60.6 deg; C = 78.6 deg

8. A = 20.7 deg; B = 127.2 deg; C = 32.1 deg

Step-by-step explanation:

Law of Cosines

$c^2 = a^2 + b^2 - 2ab \cos C$

You know the lengths of the sides, so you know a, b, and c. You can use the law of cosines to find C, the measure of angle C.

Then you can use the law of cosines again for each of the other angles. An easier way to solve for angles A and B is, after solving for C with the law of cosines, solve for either A or B with the law of sines and solve for the last angle by the fact that the sum of the measures of the angles of a triangle is 180 deg.

7.

We use the law of cosines to find C.

$18^2 = 12^2 + 16^2 - 2(12)(16) \cos C$

$324 = 144 + 256 - 384 \cos C$

$-384 \cos C = -76$

$\cos C = 0.2$

$C = \cos^{-1} 0.2$

$C = 78.6^\circ$

Now we use the law of sines to find angle A.

Law of Sines

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

We know c and C. We can solve for a.

$\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$

$\dfrac{12}{\sin A} = \dfrac{18}{\sin 78.6^\circ}$

Cross multiply.

$18 \sin A = 12 \sin 78.6^\circ$

$\sin A = \dfrac{12 \sin 78.6^\circ}{18}$

$\sin A = 0.6535$

$A = \sin^{-1} 0.6535$

$A = 40.8^\circ$

To find B, we use

m<A + m<B + m<C = 180

40.8 + m<B + 78.6 = 180

m<B = 60.6 deg

8.

I'll use the law of cosines 3 times here to solve for all the angles.

Law of Cosines

$a^2 = b^2 + c^2 - 2bc \cos A$

$b^2 = a^2 + c^2 - 2ac \cos B$

$c^2 = a^2 + b^2 - 2ab \cos C$

Find angle A:

$a^2 = b^2 + c^2 - 2bc \cos A$

$8^2 = 18^2 + 12^2 - 2(18)(12) \cos A$

$64 = 468 - 432 \cos A$

$\cos A = 0.9352$

$A = 20.7^\circ$

Find angle B:

$b^2 = a^2 + c^2 - 2ac \cos B$

$18^2 = 8^2 + 12^2 - 2(8)(12) \cos B$

$324 = 208 - 192 \cos A$

$\cos B = -0.6042$

$B = 127.2^\circ$

Find angle C:

$c^2 = a^2 + b^2 - 2ab \cos C$

$12^2 = 8^2 + 18^2 - 2(8)(18) \cos B$

$144 = 388 - 288 \cos A$

$\cos C = 0.8472$

$C = 32.1^\circ$

8 0

2 years ago

Caroline bought 20 shares of stock at 10 1/2, and after 10 months the value of the stocks was 11 1/4. If Caroline were to sell a

mojhsa [17]

11.25 - 10.5 = 0.75

Knowing that this is the profit she makes with one share, so:

20 * 0.75 = 15

Caroline made $15 profit.

7 0

3 years ago

Explain why you would make one of the addends a tens number

saveliy_v [14]

What does that mean?

7 0

3 years ago

Any body know the answers to these

snow_lady [41]

For 3. he answer is a
for 4 the answer is two

6 0

3 years ago

Find the equation of the line through (6,8) which is parallel to the line y=4x−6.

Ratling [72]

Answer:

$y=4x-16$

Step-by-step explanation:

Hi there!

<u>What we need to know:</u>

Linear equations are typically organized in slope-intercept form: $y=mx+b$ where m is the slope of the line and b is the y-intercept (the value of y when x is 0)
Parallel lines always have the same slope

<u>1) Determine the slope (m)</u>

<u /> $y=4x-6$ <u />

4 is in the place of m, making it the slope. Because parallel lines have the same slope, the slope of the line is therefore 4. Plug this into $y=mx+b$ :

$y=4x+b$

<u>2) Determine the y-intercept (b)</u>

$y=4x+b$

Plug in the given point (6,8) and solve for b

$8=4(6)+b\\8=24+b$

Subtract 24 from both sides to isolate b

$8-24=24+b-24\\-16=b$

Therefore, the y-intercept of the line is -16. Plug this back into $y=4x+b$ :

$y=4x-16$

I hope this helps!

3 0

3 years ago