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svetlana [45]
4 years ago
15

Plz help ill give you brainlist

Mathematics
1 answer:
Daniel [21]4 years ago
5 0
Consider this option:
1. formula of perimeter is: P=2(a+b), where a & b - the sides of rectangle.
2. according to the condition 2(x+(x+4))<120; ⇒ x<28
answer: D. x<28
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Is the following shape a rectangle? How do you know?
xenn [34]

The answer is C. No, the adjacent sides are not perpendicular. (Apex)

3 0
3 years ago
Read 2 more answers
The longer leg of a 30° 60° 90° triangle is 16 times square root of three how long is a shorter leg
Morgarella [4.7K]

Answer:

16

Explanation:

In your parent 30 60 90 triangle, you will see that the larger leg is x times the square root of three while the shorter leg is just x, that means you would divide the longer leg by the square root of three which gives you 16 in this question.

5 0
4 years ago
Please help. Offering all the points I have.
Simora [160]

The most specific name of the quadrilateral that's is drawn in the picture is a rhombus.

<h3>What is a rhombus?</h3>

It should be noted that a rhombus simony means a quadrilateral that has four sides and the total angles are 360°.

Also, from the information given, it can be depicted that the lengths are congruent and that the opposite sides are parallel.

Learn more about rhombus on:

brainly.com/question/20627264

#SPJ1

8 0
2 years ago
<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B%282x%2B1%29%28x-5%29%7D%7B%28x-5%29%28x%2B4%29%5E%7B2%7D%20%7D" id
dybincka [34]

i) The given function is

f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}

The domain is

(x-5)(x+4)^2\ne 0

(x-5)\ne0,(x+4)^2\ne 0

x\ne5,x\ne -4

ii) For vertical asymptotes, we simplify the function to get;

f(x)=\frac{(2x+1)}{(x+4)^2}

The vertical asymptote occurs at

(x+4)^2=0

x=-4

iii) The roots are the x-intercepts of the reduced fraction.

Equate the numerator of the reduced fraction to zero.

2x+1=0

2x=-1

x=-\frac{1}{2}

iv) To find the y-intercept, we substitute x=0 into the reduced fraction.

f(0)=\frac{(2(0)+1)}{(0+4)^2}

f(0)=\frac{(1)}{(4)^2}

f(0)=\frac{1}{16}

v) The horizontal asymptote is given by;

lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0

The horizontal asymptote is y=0.

vi) The function has a hole at x-5=0.

Thus at x=5.

This is the factor common to both the numerator and the denominator.

vii) The function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

6 0
3 years ago
Does anyones know these ?
garik1379 [7]

No I don't know your thing

5 0
3 years ago
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