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4 years ago

9

If number eight is underlined what is the position

1 answer:

pshichka [43]4 years ago

5 0

Is there any numbers to answer the question?

You might be interested in

840,000 rounded to the nearest hundred thousand

ad-work [718]

840,000 is closer to 800,000 than 900,00. Therefore the answer is 800,000

3 0

3 years ago

Read 2 more answers

Find the area. The figure is not drawn to scale

poizon [28]

Answer:

96 square inches.

Step-by-step explanation:

From given picture we see that

BC=8 in

AD=10 in

Given that

AD=AB

Then AB=10 in

Apply Pythagorean theorem in triangle ABC to find AC

$(hypotenuse)^2=(base)^2+(perpendicular)^2$

$(AB)^2=(BC)^2+(AC)^2$

$(10)^2=(8)^2+(AC)^2$

$100=64+(AC)^2$

$100-64=(AC)^2$

$36=(AC)^2$

take square root

$6=AC$

Then area of triangle ABC $=\frac{1}{2}\left(base\right)\left(altitude\right)$

$=\frac{1}{2}\left(8\right)\left(6\right)=24$

There are total 4 congruent triangles.

Then total area of the given figure = 4(24)=96 square inches.

7 0

3 years ago

If sinx = p and cosx = 4, work out the following forms :<br><br><br>

Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16}$

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

$\boxed{\sin(-x)=-\sin(x)}$

$\boxed{\cos(-x)=\cos(x)}$

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

$\sin(x)=p$

$\cos(x)=4$

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]$

Using the sum and difference identities:

$\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}$

$\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}$

$\sin(\pi -x) = \sin(x)$

$\sin(2\pi +x)=\sin(x)$

$\cos(2\pi-x)=\cos(x)$

Therefore,

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]$

$\implies [p+4] \cdot [p \cdot 4]=4p^2+16p$

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16}$

8 0

3 years ago

If a=2 b=1 find the value of a square minus b square

Usimov [2.4K]

<em>The answer is 3</em>

a= 2, b=1,

a²-b² = 2²-1²= 4-1 =3

6 0

3 years ago

Olivia and Isabella each earn d dollars for each dog she walks. One day, Olivia walks 6 dogs and gets $8 in tips. That same day,

marissa [1.9K]

Answer:

Yes they are equivalent, both equations equals $15d+20$

Step-by-step explanation:

We first simplify Olivia's equation

$6d + 8 + 9d + 12$

1. we combine like terms in the equation: $6d + 8 + 9d + 12$

$6d+9d$

$8+12$

$15d+20$

Now we simplify Isabella's

$5(3d + 4)$

1. we do distributive property

$5*3d$

$5*4$

$15d+20$

If you have any quesitons feel free to ask

8 0

3 years ago