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2 years ago

Please Help ASAP!!! Would you be able to prove this is a right triangle or would it just be a scalene triangle?

Mathematics

2 answers:

daser333 [38]2 years ago

8 0

Triange ADB is an isosceles triangle being bisected, so Triangle ADE is a right triangle

Dvinal [7]2 years ago

3 0

It is a right Right Triangle because, they have two equal sides.

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At work, Brett must check and record the internal temperature of the freezer on an hourly basis. When working properly, the temp

antoniya [11.8K]

Answer:

B. zero

Step-by-step explanation:

If the temperature is supposed to remain constant over time (the same) when working properly, then this means that there is no increase or decrease over time.

If there were a line to represent this, then it would be a straight line with a slope of 0 because the temperature would remain the same.

4 0

3 years ago

A rectangle has a length of 3x inches. The width of the rectangle is (2x+6) inches more than the length. A) Find the perimeter o

s2008m [1.1K]

Perimeter = 2*(length +width)
perimeter = 2*(3x + (3x +(2x+6)))
.. = 2*(3x +3x +2x +6)
.. = 2*(8x +6)
perimeter = 16x +12

If x=3, the perimeter is 16*3 +12 = 60 inches.

7 0

2 years ago

Help? please.........

UkoKoshka [18]

Answer: 60°

Step-by-step explanation: A straight angle is 180°

180°- 120° = 60°

8 0

2 years ago

Help me on this problem please!

nexus9112 [7]

The answer is 'F'.......

4 0

3 years ago

Read 2 more answers

hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and

Mariulka [41]

Given:

The equation is,

$2\log _3x-\log _3(x-2)=2$

Explanation:

Simplify the equation by using logarthimic property.

$\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{ \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}$

Simplify further.

$\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}$

Solve the quadratic equation for x.

$\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}$

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

$\begin{gathered} x-6=0 \\ x=6 \end{gathered}$

For (x - 3) = 0,

$\begin{gathered} x-3=0 \\ x=3 \end{gathered}$

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

$\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}$

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

$\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}$

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0

1 year ago