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olga2289 [7]
4 years ago
6

The radioisotope Phosphorus-32 is used in leukemia therapy. The half-life of this isotope is 14.26 days.

Mathematics
2 answers:
disa [49]4 years ago
8 0
B is the best answer that i can dee
Crank4 years ago
5 0
B I BELIEVE ! PLEASE LMK IF IM WRONG
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Which of the following is an example of an open equation? 5x + 7 = 12 , 12 + (-8) = 4 , 25 = 5 (5) , 8 + 0 =8
spayn [35]

An open equation is an equation that contain one or more unknow variables.

And that/those variable(s) can be found by solving the equation that make the equation true.

A variable is an English alphabet.

For example, a,b, x, y,z etc.

A variable represents a unknow value for a particular quantity.

In the given options 5x + 7 = 12 , 12 + (-8) = 4 , 25 = 5 (5) , 8 + 0 =8 .

In first option, we have unknow variable x is there.

But in all other options we don't value any variable.

Therefore, all other options can't be open equation.

So, final answer is 5x + 7 = 12 is the only open equation in given options.

7 0
3 years ago
1. A point on the ground is 50 feet from my house. The angle of elevation to the top of the house is 48 Find the height of the h
Georgia [21]

Answer:

(1)55.5 feet

(2)7.5 feet

Step-by-step explanation:

<u>Question 1</u>

The diagram is attached below.

Using Trigonometric ratio

\tan 48^\circ = \dfrac{h}{50} \\h=50 \times \tan 48^\circ\\h=55.5$ feet (to the nearest tenth)

The height of the house is 55.5 feet to the nearest tenth.

<u>Question 2</u>

In the diagram, we are required to find the distance the bird must fly to be directly above the fish. This has been represented by x.

Using Trigonometric ratio

\tan 15^\circ = \dfrac{2}{x} \\x \tan 15^\circ=2\\x=2 \div \tan 15^\circ\\x=7.5 $ feet (to the nearest tenth)

The distance the bird must fly to be directly above the fish is 7.5 feet.

6 0
3 years ago
Maurice and Johanna have appreciated the help you have provided them and their company Pythgo-grass. They have decided to let yo
Colt1911 [192]
<span><span>1.A triangular section of a lawn will be converted to river rock instead of grass. Maurice insists that the only way to find a missing side length is to use the Law of Cosines. Johanna exclaims that only the Law of Sines will be useful. Describe a scenario where Maurice is correct, a scenario where Johanna is correct, and a scenario where both laws are able to be used. Use complete sentences and example measurements when necessary.
</span>
The Law of Cosines is always preferable when there's a choice.  There will be two triangle angles (between 0 and 180 degrees) that share the same sine (supplementary angles) but the value of the cosine uniquely determines a triangle angle.

To find a missing side, we use the Law of Cosines when we know two sides and their included angle.   We use the Law of Sines when we know another side and all the triangle angles.  (We only need to know two of three to know all three, because they add to 180.  There are only two degrees of freedom, to answer a different question I just did.

<span>2.An archway will be constructed over a walkway. A piece of wood will need to be curved to match a parabola. Explain to Maurice how to find the equation of the parabola given the focal point and the directrix.
</span>
We'll use the standard parabola, oriented in the usual way.  In that case the directrix is a line y=k and the focus is a point (p,q).

The points (x,y) on the parabola are equidistant from the line to the point.  Since the distances are equal so are the squared distances.

The squared distance from (x,y) to the line y=k is </span>(y-k)^2
<span>
The squared distance from (x,y) to (p,q) is </span>(x-p)^2+(y-q)^2.<span>
These are equal in a parabola:

</span>
(y-k)^2 =(x-p)^2+(y-q)^2<span>

</span>y^2-2ky + k^2 =(x-p)^2+y^2-2qy + q^2

y^2-2ky + k^2 =(x-p)^2 + y^2 - 2qy+ q^2

2(q-k)y =(x-p)^2+ q^2-k^2

y = \dfrac{1}{2(q-k)} ( (x-p)^2+ q^2-k^2)

Gotta go; more later if I can.

<span>3.There are two fruit trees located at (3,0) and (−3, 0) in the backyard plan. Maurice wants to use these two fruit trees as the focal points for an elliptical flowerbed. Johanna wants to use these two fruit trees as the focal points for some hyperbolic flowerbeds. Create the location of two vertices on the y-axis. Show your work creating the equations for both the horizontal ellipse and the horizontal hyperbola. Include the graph of both equations and the focal points on the same coordinate plane.

4.A pipe needs to run from a water main, tangent to a circular fish pond. On a coordinate plane, construct the circular fishpond, the point to represent the location of the water main connection, and all other pieces needed to construct the tangent pipe. Submit your graph. You may do this by hand, using a compass and straight edge, or by using a graphing software program.

5.Two pillars have been delivered for the support of a shade structure in the backyard. They are both ten feet tall and the cross-sections​ of each pillar have the same area. Explain how you know these pillars have the same volume without knowing whether the pillars are the same shape.</span>
5 0
4 years ago
If (x − 2) ∶ 5 = 7 ∶ 9, then find the value of x.
WARRIOR [948]

Answer:

x=53/9

Step-by-step explanation:

(x-2):5=7:9

9x-18=35

9x=35+18

9x=53

x=53/9

4 0
3 years ago
Can anyone solve this for me???
Elena-2011 [213]

Answer:

Step-by-step explanation:

It x

5 0
3 years ago
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