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Ira Lisetskai [31]
3 years ago
10

In a triangle, the measure of the second angle is twice the measure of the first angle. The third angle is equal to the sum of t

he other angles.
Which of the following answers could represent the measures of the three angles?

A) x, 2x, 4x
B) x, 2x, 3x
C) x, 2x, 2x
Mathematics
2 answers:
attashe74 [19]3 years ago
7 0
The answer would be x,2x,3x because x is the first angle, the second angle is two times the amount (2x) and the third angle is the sum of the other two angles (x + 2x = 3x).

Hope this helps.
bija089 [108]3 years ago
3 0
X, 2x, and (x+2x)=3x

6x=180

x=30

30, 60, 90 triangle.
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Find the limit , picture provided
V125BC [204]

Answer:

d. does not exist

Step-by-step explanation:

The given limits are;

\lim_{x \to 4} f(x) =5, \lim_{x \to 4} g(x) =0 and \lim_{x \to 4} h(x) =-2

We want to find

\lim_{x \to 4} \frac{f}{g}(x)= \lim_{x \to 4} \frac{f(x)}{g(x)}

By the properties of limits, we have;

\lim_{x \to 4} \frac{f}{g}(x)= \frac{\lim_{x \to 4} f(x)}{\lim_{x \to 4} g(x)}

This gives us;

\lim_{x \to 4} \frac{f}{g}(x)= \frac{5}{0}

Division by zero is not possible. Therefore the limit does not exist.

7 0
3 years ago
What is AC?
Maru [420]
C. 6

Exploration: i got it right.
3 0
2 years ago
How to answer this question??? <br>thanks
vesna_86 [32]
\dfrac{2m}{5}+3=7(4m-3)\ \ \ \ |use\ distributive\ property:a(b-c)=ab-ac\\\\\dfrac{2m}{5}+3=28m-21\ \ \ \ \ |multiply\ both\ sides\ by\ 5\\\\2m+15=140m-105\ \ \ \ |subtract\ 140m\ from\ both\ sides\\\\-138m+15=-105\ \ \ \ |subtract\ 15\ from\ both\ sides\\\\-138m=-120\ \ \ \ |divide\ both\ sides\ by\ (-138)\\\\m=\dfrac{120:6}{138:6}\\\\\boxed{m=\frac{20}{23}}
7 0
3 years ago
Please help . I’ll mark you as brainliest if correct!!!!!!
STatiana [176]

Answer:

3 are mushed

Step-by-step explanation:

You have 12 tomatoes

All but 9 get mushed

Total tomatoes = mushed + non mushed

12 = mushed +9

Subtract 9 from each side

12-9 = mushed+9-9

12-9 = mushed

3 = mushed

7 0
3 years ago
Read 2 more answers
The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of days an
solniwko [45]

Answer:

(a) 283 days

(b) 248 days

Step-by-step explanation:

The complete question is:

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. ​(a) What is the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths? ​(b) What is the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths?

Solution:

The random variable <em>X</em> can be defined as the pregnancy length in days.

Then, from the provided information X\sim N(\mu=268, \sigma^{2}=12^{2}).

(a)

The minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths implies that:

P (X > x) = 0.11

⇒ P (Z > z) = 0.11

⇒ <em>z</em> = 1.23

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\1.23=\frac{x-268}{12}\\\\x=268+(12\times 1.23)\\\\x=282.76\\\\x\approx 283

Thus, the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths is 283 days.

(b)

The maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths implies that:

P (X < x) = 0.05

⇒ P (Z < z) = 0.05

⇒ <em>z</em> = -1.645

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\-1.645=\frac{x-268}{12}\\\\x=268-(12\times 1.645)\\\\x=248.26\\\\x\approx 248

Thus, the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths is 248 days.

8 0
3 years ago
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