All categories

4 years ago

Can you guys and gals help me

Mathematics

1 answer:

professor190 [17]4 years ago

7 0

Answer:

1. B 2. B 3. D 4. A 5. A 6. A 7. B 8. C

Step-by-step explanation:

1. 6 and 8 are right next to each other. Making them 'adjacent angles' as 'adjacent' means 'next to'

2. 2 and 3 are vertically opposite angles, and these angles are equal to each other

3. 3 and 6 are alternate angles. These angles form a 'Z' shape

4. Similar to q2, 1 and 4 are vertically opposite angles

5. Yes they are similar because there are two angles which equal

6. The angles of the big triangle will correlate to the small triangle, therefore x is 40

7. Similar triangles will have the same angles. The top angle of the small triangle is 60, as stated in the big triangle, and 70 is given. The sum of interior angles of a triangle is 180. So, add 60 and 70, then minus from 180

8. Again, same concept as q7, the angles are given to you and you just need to correspond each angle to its respective area, then minus the total of the two given angles from 180

Hope this helped you!!

You might be interested in

Which is true about constructing a bisector of the angle?

IgorLugansk [536]

When you bisect something you are cutting it in half. so in the image above when you cut it in half you would be forming 2 angles. Your answer is D. because when you cut in half you are going to have equal sides. Its like taking an 75 degrees and bisecting it and each of the new angles would be 37.5.

4 0

3 years ago

Find the angle of this sector. on the image

KatRina [158]

Answer:

$\alpha =107.4^{o}$

Step-by-step explanation:

$15=\frac{\pi (8)\alpha }{180}$

$\alpha =\frac{(15)(180)}{8\pi } =\frac{2700}{25.13}=107.4$

Hope this helps

5 0

3 years ago

Read 2 more answers

There were 94172 people at a footbsll game.

Karo-lina-s [1.5K]

WHat else is to the question?!?

4 0

4 years ago

The marked sides have equal lengths. What is the value of x?

Sholpan [36]

60-12=4x
48=4x
x=12

Hope this helps.

6 0

4 years ago

Read 2 more answers

Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com

fredd [130]

$f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}$

$f$ has critical points where the derivative is 0:

$2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1$

The second derivative is

$f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}$

and $f''(-1)=\frac2e>0$ , which indicates a local minimum at $x=-1$ with a value of $f(-1)=\frac1e$ .

At the endpoints of [-2, 2], we have $f(-2)=1$ and $f(2)=e^8$ , so that $f$ has an absolute minimum of $\frac1e$ and an absolute maximum of $e^8$ on [-2, 2].

So we have

$\dfrac1e\le f(x)\le e^8$

$\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx$

$\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}$

5 0

3 years ago