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uysha [10]
3 years ago
15

635-521+159-3 what is the answer?

Mathematics
2 answers:
svlad2 [7]3 years ago
7 0
The answer is 270. Brainliest answer?
Novay_Z [31]3 years ago
3 0
The answer is 270. 

I hope this is helpful! :D
You might be interested in
What is f(x) = 8x2 4x written in vertex form? f(x) = 8(x one-quarter) squared – one-half f(x) = 8(x one-quarter) squared – one-s
Maksim231197 [3]

The vertex form of the quadratic equation, for the vertex points at which the parabola crosses its symmetry is,

f(x)=8[(x+\dfrac{1}{4})^2]-\dfrac{1}{2}

<h3>What is vertex form of parabola?</h3>

Vertex form of parabola is the equation form of quadratic equation which is used to find the coordinate of vertex points at which the parabola crosses its symmetry.

The standard equation of the vertex form of parabola is given as,

y=a(x-h)^2+k

Here, (<em>h, k</em>) is the vertex point.

The given quadratic equation in the problem is,

f(x)=8x^2+4x

To isolate the term x^2 , take out the common number 8 from the equation,

f(x)=8(x^2+0.5x)\\

Add and subtract 0.5/2 in the equation as,

f(x)=8(x^2+0.5x+\dfrac{0.5}{2}-\dfrac{0.5}{2})\\f(x)=8(x^2+0.5x+\dfrac{1}{16}-\dfrac{1}{16})\\f(x)=8[(x+0.25)^2-\dfrac{1}{16}]\\f(x)=8[(x+\dfrac{1}{4})^2]-\dfrac{1}{2}

The above equation is vertex form of the given quadratic equation. Thus, the option A is the correct option.

Learn more about the vertex form of the parabola here;

brainly.com/question/17987697

6 0
2 years ago
How would you write 1 3/16 as a decimal
galben [10]
1.1875................
8 0
4 years ago
Read 2 more answers
What expression can be used for estimating 868 divided by 28?<br> Pls help me<br> Thanks
horrorfan [7]
868 is almost 870. 28 is almost 30.

870 / 30 = 29

So 868 / 28 is approximately 29. 
7 0
3 years ago
What is the volume of the prism?
KatRina [158]

Answer:

D.96ft^3......is the and...hope it helps

3 0
3 years ago
Is the triangle obtuse, acute, equilateral or right?
Stells [14]

9514 1404 393

Answer:

  obtuse

Step-by-step explanation:

The law of cosines tells you ...

  b² = a² +c² -2ac·cos(B)

Substituting for a²+c² using the given equation, we have ...

  b² = b²·cos(B)² -2ac·cos(B)

We can subtract b² to get a quadratic in standard form for cos(B).

  b²·cos(B)² -2ac·cos(B) -b² = 0

Solving this using the quadratic formula gives ...

  \cos(B)=\dfrac{-(-2ac)\pm\sqrt{(-2ac)^2-4(b^2)(-b^2)}}{2b^2}\\\\\cos(B)=\dfrac{ac}{b^2}\pm\sqrt{\left(\dfrac{ac}{b^2}\right)^2+1}

The fraction ac/b² is always positive, so the term on the right (the square root) is always greater than 1. The value of cos(B) cannot be greater than 1, so the only viable value for cos(B) is ...

  \cos(B)=\dfrac{ac}{b^2}-\sqrt{\left(\dfrac{ac}{b^2}\right)^2+1}

The value of the radical is necessarily greater than ac/b², so cos(B) is necessarily negative. When cos(B) < 0, B > 90°. The triangle is obtuse.

4 0
3 years ago
Read 2 more answers
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