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kupik [55]
2 years ago
7

Numbers expressed using exponents are called .

Mathematics
2 answers:
Luden [163]2 years ago
5 0

Answer: Powers

Step-by-step explanation:

Numbers expressed using exponents are called powers

N76 [4]2 years ago
4 0

Answer:

numbers expressed using exponents are called powers

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Reflect shape A in the line y = x.
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8 0
2 years ago
How are fractions divided the right way?
Lady_Fox [76]

You would keep the first fraction the same , turn the division sign into a multiplication sign , and flip the 2nd fraction upside down.


10/14 ÷ 7/16


10/14 x 16/7= 80/49 = 1 31/49

8 0
3 years ago
Logan wants to move to the city
Usimov [2.4K]

Answer:good for him

Step-by-step explanation:

7 0
3 years ago
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A bad punter on a football team kicks a football approximately straight upward with an initial velocity of 89 ft/sec.
Andrej [43]

Answer:

See below

Step-by-step explanation:

Vertical velocity will be affected by gravity in this scenario

df = do + vo t + 1/2 a t^2         do = original height = 4 ft      a = -32.2 ft/s^2

<u>df = 4 + 89 t  - 1/2 (32.2) t^2        df = height</u>

<u />

On the way up and the way down, the ball may reach height of 102.2125 ft :

102.2125 = 4 + 89 t - 1/2 (32.2) t2   re-arrange to:

-16.1 t^2 + 89t - 98.2125 =0  

      Use Quadratic Formula to find <u>t =   1.5 and 4.0 s </u>

6 0
2 years ago
F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa
DerKrebs [107]

a. Parameterize C by

\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath

with 0\le t\le1.

b/c. The line integral of \vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath over C is

\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt

=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt

=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt

=\displaystyle10\int_0^1\mathrm dt=\boxed{10}

d. Notice that we can write the line integral as

\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)

By Green's theorem, the line integral is equivalent to

\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy

where D is the triangle bounded by C, and this integral is simply twice the area of D. D is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0
3 years ago
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