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baherus [9]
3 years ago
8

NO LINKS OR FILES!!!!!!! How many millimeters is this line segment?

Mathematics
1 answer:
34kurt3 years ago
7 0

Answer:

remember this line it will help you

<h3>Milli centi deci meter dega hecto kilometer</h3>

Step-by-step explanation:

<h2>milli to centi takes 10 points difference so option c. 40 millimeters is the correct answer</h2>
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Solve for x and y in the system of equation<br>3x²+4y=17 <br>2x²+5y=2<br>I'll mark brainliest​
Novay_Z [31]

Answer:

x=\pm \sqrt{11} \\y=-4.

Step-by-step explanation:

\left \{ {{3x^2+4y=17} \atop {2x^2+5y=2}} \right. \\\\Then\\\\\left \{ {{2 \times (3x^2+4y)=2 \times 17} \atop {3\times(2x^2+5y)=3\times2}} \right. \\\\\left \{ {{6x^2+8y=34 \quad (1)} \atop {6x^2+15y=6\quad (2)}} \right.\\\\(1)-(2) \Rightarrow -7y=28 \Rightarrow y =-4.\\Since \: \:  2x^2+5y=2 \Rightarrow 2x^2=2-5y\\\Rightarrow 2x^2=2+20=22\\\Rightarrow x^2=22:2=11\\\Rightarrow x=\pm \sqrt{11}.

7 0
3 years ago
If f(x)=4^x, find f(4)
lina2011 [118]

Answer:

\huge\boxed{f(4) = 256}

Step-by-step explanation:

<u>Given function is:</u>

f(x) = 4^x

Put x = 4

So,

f(4) = 4^4\\\\f(4) = 256\\\\\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
6 0
3 years ago
How many diagonals can be drawn from one vertex of a hexagon
Dmitrij [34]
Three diagonals can be drawn in a hexagon mmyyy dude.
5 0
4 years ago
Read 2 more answers
Biology application. The approximate number of calories C that an animal needs each day is given by
Reika [66]

Answer:

C = 288 kg

Step-by-step explanation:

Given:

C = 72m^2/4

where,

C = approximate number of calories

m = animal mass in kilograms.

Find C when m = 16

C = 72m^2/4

= 72 * (16)^2/4

= 72 * (16)^1/2

= 72 * √16

= 72 * 4

= 288

C = 288 kg

7 0
3 years ago
maria has a total of 95 dimes and quarters. If the total value of the coins is $23.45, how many quarters does she have? Please S
NARA [144]
This is the concept of application of quadratic equations; To get the number of dimes and quarters we proceed as follows;
suppose there are x dimes and y quarters;
x+y=95.......i
but we know;
$0.1=1 dimes
$0.25=1 quarters 
thus
0.1x+0.25y=23.45.....ii
solving equation i and ii by substitution we shall have:
from i;
x=95-y
thus substituting the value of x in equation ii we get
0.1(95-y)+0.25y=23.45
9.5-0.1y+0.25y=23.45

collecting like terms we get:
0.15y=13.95
dividing both sides by 0.15 we get;
y=13.95/0.15=93
x=95-93=2
therefore we conclude that there were 2 dimes and 93 quarters

4 0
3 years ago
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