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ICE Princess25 [194]
3 years ago
5

What is the area of the triangle? ​

Mathematics
2 answers:
Komok [63]3 years ago
8 0

Answer:

Its the second option. 742.5

netineya [11]3 years ago
8 0

Answer:

A = 742.5in^2

Step-by-step explanation:

When solving the area of a triangle we need to have A= hb/2

h: being the height of the triangle has and b: being the base of the triangle when joining these two number you always have divide by 2 when finding an area of a triangle giving you your final equation:

A = hb/2 = 33 x 45/ 2 = 742.5

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A pot contains 25 ounces of soil. You use 4 1/4
Wewaii [24]

Answer: multiply 4 1/4 by 5 then see if you have enough .

Step-by-step explanation: multiply 4 1/4 by 5 because you have used 1 herb already so if you do that you could have enough in there to pot 5 .

6 0
3 years ago
1+1=?
miv72 [106K]

Answer:

11

Step-by-step explanation:

one plus another one is 11

0+1+0+1 = 11

4 0
3 years ago
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Why does a y-intercept not count as a zero?
Genrish500 [490]

Answer:

It only counts as a zero when the y-intercept is (0,0).

Step-by-step explanation:

The zeros of a quadratic function are always written as (x,0), while the y-intercept is always written as (0,y). Therefore, in order for a y-intercept to be a zero, it must be (0,0), because the y-coordinate in any zero is 0. At any other time, the y-intercept is not a zero.

7 0
4 years ago
180 lb 12 oz - 79.3 kg = ___________ kg
Brrunno [24]

Answer:

180lb\ 12 oz - 79.3kg = 2.8kg

Step-by-step explanation:

Given

180\ lb 12\ oz - 79.3kg

Required

Solve

First, convert 180lb to kg

1lb \to 0.454kg

180lb \to 180 *0.454kg

180lb \to 81.7kg -- to 1dp

Next, convert 12oz to kg

1oz \to 0.03kg

12oz \to 12 * 0.03kg

12oz \to 0.4kg --- to 1 dp

So, the expression becomes:

180\ lb 12\ oz - 79.3kg

81.7kg + 0.4kg - 79.3kg

= 2.8kg

Hence:

180lb\ 12 oz - 79.3kg = 2.8kg

3 0
3 years ago
Find e^cos(2+3i) as a complex number expressed in Cartesian form.
ozzi

Answer:

The complex number e^{\cos(2+31)} = \exp(\cos(2+3i)) has Cartesian form

\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2).

Step-by-step explanation:

First, we need to recall the definition of \cos z when z is a complex number:

\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}.

Then,

\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}. (I)

Now, recall the definition of the complex exponential:

e^{z}=e^{x+iy} = e^x(\cos y +i\sin y).

So,

e^{2i-3} = e^{-3}(\cos 2+i\sin 2)

e^{-2i+3} = e^{3}(\cos 2-i\sin 2) (we use that \sin(-y)=-\sin y).

Thus,

e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)

Now we group conveniently in the above expression:

e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2.

Now, substituting this equality in (I) we get

\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2.

Thus,

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right].

5 0
3 years ago
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