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uysha [10]
2 years ago
6

Grace drew a triangle. It's sides were 12 mm, 10 mm, and 11 mm. It has one obtuse angle and two acute angles. What triangle did

she draw?
Mathematics
1 answer:
Mariana [72]2 years ago
4 0

Answer:

Scalene triangle

Step-by-step explanation:

scalene triangle is a triangle where the lengths of all three sides are different, it comprises of both obtuse and acute angles,all giving a sum of 180 degrees

You might be interested in
Point A is the midpoint of side XZ and point B is the
Katyanochek1 [597]

Answer:

Option (2)

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Point A is the midpoint of side XZ and point B is the  midpoint of side YZ.

Triangle XYZ is cut by line segment AB. Point A is the midpoint of side XZ and point B is the midpoint of side YZ. The length of XY is (5x - 7), the length of AB is (x + 1), and the lengths of XA and ZA are (2x - 2). The lengths of YB and BZ are congruent.

What is AX ?

2 units

4 units

6 units

8 units

From the figure attached,

XYZ is a triangle having A and B as the midpoints of the sides XZ and YZ.

By the theorem of midpoints,

AB = \frac{1}{2}\text{(XY)}

Since AB = (x + 1) and XY = (5x - 7)

(x + 1) = \frac{1}{2}(5x-7)

2x + 2 = 5x - 7

5x - 2x = 2 + 7

3x = 9

x = 3

Therefore, side AB = 3 + 1 = 4 units

Side XY = (5x - 7)

             = 15 - 7

             = 8 units

Side XA = 2(x - 1)

              = 4 units

Therefore, Option (2) will be the answer.

6 0
2 years ago
The 6 members of the homecoming decorating committee want to make 525 paper flowers for the homecoming dance. Each flower takes
stiv31 [10]
<h3>Answer:</h3>

6 hours 36 minutes, or 6 hours 31 1/2 minutes

<h3>Step-by-step explanation:</h3>

Dividing the task among the 6 committee members means each member will be making 525/6 = 87.5, that is, either 87 or 88 flowers. (We doubt that two people working together on the same flower will finish it in half the time.)

Each of the three committee members making 88 flowers will take ...

... 88 × 4 1/2 minutes = 396 minutes = 6 hours 36 minutes

Each member making 87 flowers will be finished 4 1/2 minutes sooner, after 6 hours 31 1/2 minutes.

_____

<em>Comment on the problem</em>

This problem seemingly invites you to divide the labor evenly among committe members. Doing that would give you an answer of 6 hours 33 3/4 minutes. You need to ask yourself whether that is practical for this situation.

4 0
2 years ago
Evaluate log4 0.25 ?
Vsevolod [243]

Answer:

-1

Step-by-step explanation:

Recall that 0.25 = 1/4, and that 1/4 = 1/[4^(-1)  Notice that 4 is the base of this log system.

Then log4 0.25 = log4(1/4) = log4(4^(-1) ) = -1 (answer)

Check:  Does 4^(-1) = 0.25?  Yes

8 0
2 years ago
A manageress earned £24,500 last year. This year she earns £25,235.
Korolek [52]

Answer:

Step-by-step explanation:

245000 last year

This year 25235

(y2 - y1) / y1)*100 = your percentage change

(where y1=start value and y2=end value)  

(( £25.235 - £24.500) / £24.500) * 100 = 0 %

There ain't no percentage change as there needs to be a bigger difference between the two numbers plus u should use the formula

7 0
1 year ago
Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

3 0
2 years ago
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