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4 years ago

Tell whether the triangle with the given side lengths is a right triangle

Mathematics

2 answers:

RSB [31]4 years ago

8 0

No it’s not a right triangle

Lapatulllka [165]4 years ago

8 0

Answer: This is not a right triangle but rather an acute triangle.

hope this helps (:

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Juana and Andy completed 10 math problems total. Juana completed 6 of the problems. Write an equation to determine the number of

kompoz [17]

Step-by-step explanation:

let the number of questions solved by Andy=x

this implies that 6+x=10

making x the subject of the equation

x=10-6

x=4

therefore Andy solved 4 questions

7 0

2 years ago

Find the measure of CD. Round to the nearest tenth.

e-lub [12.9K]

Hi, I explained it here. Hope this helps :)

4 0

3 years ago

Read 2 more answers

Which constant number would be an to both sides of the equation rn order to complete the square for the quadratic function 1=x^2

AnnZ [28]

Answer:

k = 9

Step-by-step explanation:

Given

$1= x^2 - 6x$

Required

The perfect number

The coefficient c of x in the above equation is:

$c = -6$

Divide by 2

$c/2 = -3$

Square both sides

$(c/2)^2 = 9$

Hence, the constant to add is:

$k = 9$

8 0

3 years ago

Find the interest on $4,500 at 5.2% for 2 and a half years. Which of the following equations would be used to solve this problem

alekssr [168]

Answer:

The answer is A=4500(1 + (0.052 × 2.5)) = 5085

Step-by-step explanation:

Since that is what you need the answer is C

4 0

4 years ago

On what intervals is the graph of f(x)=x^3-2x^2+5x+1 increasing?

amid [387]

$f(x)=x^3-2x^2+5x+1\\ f'(x)=3x^2-4x+5\\
3x^2-4x+5=0\\\Delta=(-4)^2-4\cdot3\cdot5=16-60=-44$
$\Delta0 \Rightarrow$ the graph of the parabola is above the x-axis, so the derivative is always positive and therefore the initial function is increasing in its whole domain.

$f(x)=0.5x^2-6\\
f'(x)=x$
The function is decreasing when its first derivative is negative. The first derivative of this function is negative for $x$ so for $x\in(-\infty,0)$ the function is decreasing.

$f(x)=\dfrac{x+1}{x-1}\qquad(x\not=1)\\
f'(x)=\dfrac{x-1-(x+1)}{(x-1)^2}=-\dfrac{2}{(x-1)^2}$
The function is increasing when its first derivative is positive. The first derivative of this function is always negative therefore this function is never increasing.

3 0

4 years ago