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3 years ago

13

Lions live in groups called prides. Lion populations would most likely fall in which category of population distribution?(A) eve

n(B) random(C) clumped(D) density

2 answers:

Ket [755]3 years ago

5 0

Answer: clumped

hope this helps

AveGali [126]3 years ago

4 0

Answer:

Its C. clumped

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Which expression is equal to -4<br><br>A - (-12/3)<br><br>B - (-20/5)<br><br>C -8/-2<br><br>D 16/4

Sonja [21]

The expression that is equal to -4 is D 16/-4 which means option D is correct. ( it should be noted that option D is not written properly)

<h3>Why is that 4 is equals to 16/-4?</h3>

In a mathematical expression when as negative vale is been used to multiply a negative value, then we will have a positive values , same thing with when a negative value is used to multiply a positive value, a negative value will be gotten.

In the case of the question, if positive value is used to divide a negative value, a negative value will be gotten, hence expression that is equal to -4 is 16/-4 which is means optionD is correct.

Therefore, option C is correct.

Learn more about mathematical expression at:

brainly.com/question/723406

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6 0

2 years ago

303/4 - -929/8 -59/2

ser-zykov [4K]

303/4 - -929/8 -59/2
= 1299/8

3 0

3 years ago

If four angles of a convex pentagon measure 120°, 93°, 108°, and 101°, find the measure of the fifth angle. Show equations and w

olga nikolaevna [1]

It is given in the question that

If four angles of a convex pentagon measure 120°, 93°, 108°, and 101°, find the measure of the fifth angle.

First we have to find the sum of measurement of all the angles of a convex pentagon, by using the formula

$=(n-2)*180$

Where n is the number of sides. And for polygon, it is 5 .SO the sum of measurement of angles of a polygon is

$=(5-2)*180 = 540 \degree$

Let the measurement of missing angle be x .

So we have

$120+93+108+101+x=540$

$422+x =540$

$x=540-422=118$

7 0

3 years ago

Log5(x-1)+log5(x+3)-1=0

jonny [76]

Answer:

there are no real solutions

Step-by-step explanation:

$\log_5(x-1)+\log_5(x+3)-1=0\\\\\log_5(x-1)+\log_5(x+3)=1$

there is a rule that says

$\log_b(a)+\log_b(c)=\log_b(ac)$

so we have

$\log_5(x-1)+\log_5(x+3)=\log_5((x-1)(x+3))=1$

and we have the definition

$\log_a(b)=c\\\\a^{c} =b$

so we have

$5^{1} = (x-1)(x+3)\\\\5=x^{2} -1x+3x-3\\\\5=x^{2} +2x-3\\\\0=x^{2} +2x-3-5\\\\0=x^{2} +2x-8\\\\$

and using the quadratic formula we get that

there are no real solutions

3 0

4 years ago

Read 2 more answers

at a local gym you can either pay $90 for a monthly pass then pay for dollars each time you go to the gym or you can pay $13 eac

vovangra [49]

Answer:1st option

Step-by-step explanation:

3 0

3 years ago