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Fynjy0 [20]
4 years ago
12

I don’t understand how do you do this I have a assignment for tommorow some one help

Mathematics
2 answers:
frosja888 [35]4 years ago
8 0

Answer:

500 ML

Step-by-step explanation:

for every 1 liter, it is 1,000ML

so 1 L / 2 is 500ML

and 500 on your beaker is located in between 600 and 400

Hope this helps :)

Veseljchak [2.6K]4 years ago
5 0
The answer would be 500 mL
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Phil is standing x+5 feet from the base of a building. The height of the building is 6x−1 feet. Find the function that the strai
Sergio [31]

Answer:

f(x)=\sqrt{37x^{2}-2x+26}

Step-by-step explanation:

we know that

To compute the required distance, using properties of triangles

Applying the Pythagoras Theorem

see the attached figure to better understand the problem

PT^{2}=(6x-1)^{2}+(x+5)^{2}

convert to expanded form

PT^{2}=36x^{2}-12x+1+x^{2}+10x+25

PT^{2}=37x^{2}-2x+26

square root both sides

PT=\sqrt{37x^{2}-2x+26}

Convert to function notation

f(x)=\sqrt{37x^{2}-2x+26}

5 0
4 years ago
The perimeter of a rectangle is 46 in and the diagonal is 17 in. Find the area of the rectangle.
OlgaM077 [116]
P=2*(L+l)=46
L+l=46:2
L+l=23
(L+l)^2=529
(L^2+l^2)+2l*L=529

D= V L^2+l^2= 17
L^2+l^2=289

so 289+2l*L=529
     2l*L=529-289=240
l*L=240:2
A=120 


5 0
3 years ago
Determine two angles between 0° and 360° that have a cosine of negative root3/2 without using a calculator
Naily [24]

Answer:

150° and 210°

Step-by-step explanation:

1) to drow a circle with radius=1 in the system of coordinates like in the attached picture;

2) to mark point (-√3/2)≈-0.866 on X-axis (in the picture it is the point M);

3) to graph line AB through point M, the line AB || Y-axis and the points A and B are on the circle with radius=1;

4) to measure the angles ∠ROA and ∠ROB (the points are marked with red colour);

5) finally, m(∠ROA)=150°; m(∠ROB)=210°.

5 0
3 years ago
Identify the type of function represented by f(x)=3(1.5)*
Brums [2.3K]

Answer:

Exponential growth

I Hope this is helpful

4 0
3 years ago
Read 2 more answers
Distance between the points
Harman [31]
To understand the distance formula, you first need to understand the Pythagorean Theorem. For a refresher, the theorem states that the square of the legs of a right triangle is equal to the the square of its hypotenuse (the side opposite the right angle), or in symbols:

a^2+b^2=c^2, where a and b are the lengths of the legs, and c is the length of the hypotenuse. In the context of the x-y plane, the legs of the triangle correspond to separate x and y values on the plane, and the hypotenuse corresponds to a straight line between two points on that plane.

To find the distance between the points you've listed, (2√5,4) and (1,2√3), we'll first need to find the "legs" of the triangle. To find the length of the x leg, we'll just need the distance between the x values of the points, which we find to be 2√5-1. We do the same for the y component, which ends up being 4-2√3. Now that we have our legs, we're ready to find the hypotenuse - or the distance.

Going back to Pythagorus's equation, we have:

(2 \sqrt{5}-1)^{2}+(4-2 \sqrt{3})^{2}=d^2

where d, the hypotenuse of the triangle, means "distance."

To solve for d, we take the square root of both sides:

d= \sqrt{(2 \sqrt{5}-1)^2+(4-2 \sqrt{3} )^2}

And from there, all that's left to do is solve the right side of the equation, which just ends up being rote calculation.

Edit: I'll go through the steps of that calculation here. We'll start by expanding each of the squared terms inside the radical:

(2 \sqrt{5}-1)^2=(2 \sqrt{5}-1)(2 \sqrt{5}-1)=(2 \sqrt{5}-1)2 \sqrt{5}-(2 \sqrt{5}-1)
=(2 \sqrt{5})^2-2 \sqrt{5}-2 \sqrt{5}+1=20-4\sqrt{5}+1

(4-2\sqrt{3})^2=(4-2\sqrt{3})(4-2\sqrt{3})=(4-2\sqrt{3})4-(4-2\sqrt{3})2\sqrt{3}
=16-8\sqrt{3}-8\sqrt{3}+(2\sqrt{3})^2=16-16\sqrt{3}+12

Putting those values back under the radical:

\sqrt{20-4\sqrt{5}+1+16-16\sqrt{3}+12}

Collecting constants:

\sqrt{49-4\sqrt{5}-16\sqrt{3}}

If you wanted an exact answer, this messy-looking thing would be it, and you can verify those results on WolframAlpha if you'd like. If you want an approximation, just enter that expression in to the online calculator of your choice, and it should give out the value of approx. <span>3.51325.</span>

In general, if you want to solve for the distance between two points (y_{1},x_{1}) and (y_{2},x_{2}), the formula is:

d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}
4 0
3 years ago
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