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barxatty [35]
3 years ago
15

PLEASE HELP! 10 Points!!

Mathematics
2 answers:
Lina20 [59]3 years ago
7 0
It’s D everything has to be 180 degrees
MAVERICK [17]3 years ago
6 0

Answer:

c185

Step-by-step explanation:

trust me m8 I'm 21 I wont steer you wrong

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Evaluate 0.00048×0.81×10×(10)-7 ÷0.027×0.04x(10)6​
alexandr402 [8]

Answer:

Step-by-step explanation:

0.00048*.81=0.0003888

0.0003888*10=0.003888

0.003888*(10)=0.3888

0.3888-7=-6.6112

-6.6112*0.027=-224.8592593

-224.8592593*0.04= -9.794370372

-9.794370372*(10)=-97.94370372

-97.94370372*6= -587.6622223

5 0
1 year ago
Which statements are true?
Rina8888 [55]

Answer:

The first one and the last one are true (Since −6 is 6 units to the left of 0, |−6|=6 and −6 is closer to zero on the number line than −7, so |−6|<|−7|.)

Step-by-step explanation:

Absolute value is the distance to zero, and it is always positive. That means the positive and negative versions of a number have the same absolute value. That means the first one is true, and the second and third ones are false. For the last one, -6 is closer to zero, so that means it would be true (The absolute value of -6 is 6, and the absolute value of -7 is 7).

4 0
2 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
2 years ago
Read 2 more answers
Write the following expression in expanded form 3(4y+x-8)
Makovka662 [10]
12y+3x-24 you times everything in the bracket by 3
3 0
2 years ago
Read 2 more answers
If f(x)= x+2<br>g(x)=x+1<br>find gof(x)​
ohaa [14]

Answer:

gof(x) = 3

Step-by-step explanation:

Given that,

f(x) = x+2 and g(x) = x+1

We need to find gof(x).

gof(x) means g[f(x)].

g[f(x)] = g(x +2)

= x+2+1

= x+3

Hence, the value of gof(x) is 3.

6 0
2 years ago
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